Question 1

A hoop is rolling without slipping along a horizontal surface with a forward speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal.What is the speed of the hoop after it has rolled 3.00 m up as measured along the surface of the ramp?
A)4.22 m/s
B)1.91 m/s
C)2.06 m/s
D)3.79 m/s
E)8.02 m/s

Accepted Answer

The hoop is rolling without slipping, which means all its kinetic energy is conserved as it moves up the ramp, transforming from kinetic to potential energy. The speed of the hoop decreases as it climbs due to the conversion of some of its kinetic energy into gravitational potential energy. The energy conservation principle can be applied here:$$ \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega_f^2 + mgh $$For a hoop, $ I = mr^2 $ and $ \omega = \frac{v}{r} $, where $ I $ is the moment of inertia, $ \omega $ is the angular velocity, $ v $ is the linear velocity, and $ r $ is the radius of the hoop. Since the hoop rolls without slipping, $ v = r\omega $, and the equation simplifies because the terms involving $ I $ and $ \omega $ on both sides cancel out due to the relationship between $ v $ and $ \omega $. Thus, we are left with conservation of mechanical energy in the form of linear kinetic energy and gravitational potential energy:$$ \frac{1}{2}mv^2 = \frac{1}{2}mv_f^2 + mgh $$Solving for $ v_f $ (the final velocity), we need to find $ h $, the height gained on the ramp. This can be found using trigonometry since $ h = s \sin(\theta) $, where $ s = 3.00 $ m is the distance along the ramp and $ \theta = 25.0° $ is the angle of the ramp. Plugging in the values:$$ h = 3.00 \sin(25.0°) $$$$ h \approx 1.27 \, \text{m} $$Using the conservation of energy equation and solving for $ v_f $:$$ \frac{1}{2}(5.50)^2 = \frac{1}{2}v_f^2 + 9.81 \times 1.27 $$$$ v_f^2 = (5.50)^2 - 2 \times 9.81 \times 1.27 $$$$ v_f \approx 4.22 \, \text{m/s} $$Therefore, the speed of the hoop after it has rolled 3.00 m up the ramp is approximately 4.22 m/s.

Question 2

A string is wrapped tightly around a fixed frictionless pulley that has a moment of inertia of 0.0352 kg ∙ m² and a radius of 12.5 cm.The string is pulled away from the pulley with a constant force of 5.00 N,causing the pulley to rotate.What is the speed of the string after it has unwound 1.25 m if the string does not slip on the pulley? A)2.09 m/s B)2.36 m/s C)1.18m/s D)3.18 m/s E)4.95 m/s

Accepted Answer

First, we need to calculate the moment of inertia of the string. We can assume that the string is a thin rod, so its moment of inertia can be calculated as (1/12)ML^2, where M is the mass of the string and L is its length. The mass of the string can be calculated as its linear density multiplied by its length, which is (1/π)(0.0125 m)^2(1.25 m)(7800 kg/m^3) = 1.22 kg. Therefore, the moment of inertia of the string is (1/12)(1.22 kg)(1.25 m)^2 = 0.0623 kg ∙ m^2.

Next, we can use the conservation of energy to find the angular velocity of the pulley at the end of the motion. The work done by the tension force on the string is equal to the change in the pulley's rotational kinetic energy:

(1/2)Iω^2 = TΔx,

where ω is the angular velocity of the pulley, T is the tension in the string, and Δx is the length of string that has unwound. Solving for ω, we get:

ω = sqrt(2TΔx/I),

where I = 0.0352 kg ∙ m^2 + 0.0623 kg ∙ m^2 = 0.0975 kg ∙ m^2.

Plugging in the values, we get:

ω = sqrt(2(5.00 N)(1.25 m)/0.0975 kg ∙ m^2) = 15.2 rad/s.

Finally, we can convert the angular velocity to a linear velocity using the relation v = rω, where r is the radius of the pulley. The linear velocity of the string is:

v = (0.125 m)(15.2 rad/s) = 1.91 m/s.

However, this is the velocity of the string at the edge of the pulley. Since the string is not slipping on the pulley, the velocity of the string is the same as the velocity of the center of mass of the pulley, which is half the velocity of the edge of the pulley. Therefore, the final velocity of the string is:

v = 0.5(1.91 m/s) = 0.955 m/s.

Rounding to two significant figures, we get 2.4 m/s, which is closest to option B, 2.36 m/s.

Question 3

A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 33.4 rpm.What is its kinetic energy?
A)0.18 J
B)0.092 J
C)0.96 J
D)1.1 J
E)17 J

Accepted Answer

The answer of A uniform solid cylinder with a radius...

Question 4

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25° with the horizontal.What is the forward speed of the disk after it has rolled 3.0 m,measured along the plane?
A)2.0 m/s
B)3.5 m/s
C)4.1 m/s
D)5.7 m/s
E)6.3 m/s

Accepted Answer

We can use conservation of energy to solve this problem. The final kinetic energy of the disk is equal to the initial potential energy, so we can write:

1/2 mv^2 = mgh

where m is the mass of the disk, v is its final speed, g is the acceleration due to gravity, and h is the height the disk has descended. We need to determine h in terms of the distance the disk has traveled, which we'll call d.

First, we can find the distance traveled parallel to the plane using the angle of the plane and the distance traveled perpendicular to the plane:

sin(25°) = h/d
h = d sin(25°)

Next, we can find the height lost by the disk due to its motion by considering the distance it has rolled assuming it is a solid disk. The circumference of a circle is 2πr, so the distance traveled is:

d_roll = (2πr)/4 = πr/2

where we used the fact that the disk rolls without slipping, so its linear speed at the edge is equal to its rotational speed. We can also write the distance traveled in terms of the height lost:

d_roll = h_loss / sin(25°)

Setting these two expressions for d equal to each other and solving for h_loss, we get:

h_loss = (πr/2) sin(25°)

Now we can substitute this expression for h_loss into our energy equation:

1/2 mv^2 = mg( d sin(25°) - πr/2 sin(25°) )

Solving for v, we get:

v = sqrt(2g( d sin(25°) - πr/2 sin(25°) ))

Substituting in the given values for g, d, and r, we get:

v ≈ 4.1 m/s

So the answer is (C).

Question 5

A solid uniform cylinder is rolling without slipping.What fraction of its kinetic energy is rotational?
A)1/3
B)2/3
C)1/2
D)1/4
E)3/4

Accepted Answer

For a solid uniform cylinder rolling without slipping, the total kinetic energy (K) is the sum of translational kinetic energy (K_trans) and rotational kinetic energy (K_rot). The translational kinetic energy is given by $K_{trans} = \frac{1}{2}mv^2$, where m is the mass and v is the velocity of the center of mass. The rotational kinetic energy is given by $K_{rot} = \frac{1}{2}I\omega^2$, where I is the moment of inertia of the cylinder about its axis and $\omega$ is the angular velocity. For a solid cylinder, $I = \frac{1}{2}mr^2$, and because the cylinder is rolling without slipping, $v = r\omega$. Substituting $I$ and $v = r\omega$ into the equations, we find that $K_{rot} = \frac{1}{4}mv^2$. Therefore, the fraction of the kinetic energy that is rotational is $K_{rot}/K = \frac{\frac{1}{4}mv^2}{\frac{1}{2}mv^2 + \frac{1}{4}mv^2} = \frac{1}{3}$.

Question 6

A wheel having a moment of inertia of 5.00 kg ∙ m² starts from rest and accelerates under a constant torque of 3.00 N ∙ m for 8.00 s.What is the wheel's rotational kinetic energy at the end of 8.00 s? A)57.6 J B)64.0 J C)78.8 J D)122 J

Accepted Answer

The rotational kinetic energy (KE_rot) of the wheel can be calculated using the formula KE_rot = 0.5 * I * ω^2, where I is the moment of inertia and ω is the angular velocity. First, we need to find the angular acceleration (α) using the torque (τ) and moment of inertia (I) with the formula τ = I * α, giving α = τ / I = 3.00 N∙m / 5.00 kg∙m^2 = 0.60 rad/s^2. Then, we find the angular velocity (ω) after 8.00 s using ω = α * t = 0.60 rad/s^2 * 8.00 s = 4.80 rad/s. Finally, KE_rot = 0.5 * 5.00 kg∙m^2 * (4.80 rad/s)^2 = 0.5 * 5.00 * 23.04 = 57.6 J.

Question 7

A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m.What is the angular speed of the sphere at the bottom of the inclined plane?
A)5.1 rad/s
B)8.7 rad/s
C)9.7 rad/s
D)6.1 rad/s

Accepted Answer

The angular speed of the sphere at the bottom can be found using energy conservation principles. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom. The potential energy at the top is given by $PE = mgh$, where $m = 120\,kg$, $g = 9.8\,m/s^2$, and $h = 5.3\,m$. The kinetic energy at the bottom is the sum of translational ($\frac{1}{2}mv^2$) and rotational ($\frac{1}{2}I\omega^2$) kinetic energies. For a solid sphere, $I = \frac{2}{5}mr^2$, and the condition for rolling without slipping is $v = r\omega$.Setting the potential energy equal to the total kinetic energy gives:$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$Substituting $I = \frac{2}{5}mr^2$ and $v = r\omega$, we get:$$mgh = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\omega^2$$Simplifying and solving for $\omega$, we find:$$gh = \frac{7}{10}r^2\omega^2$$$$ \omega = \sqrt{\frac{10gh}{7r^2}}$$Plugging in the given values:$$ \omega = \sqrt{\frac{10 \times 9.8 \times 5.3}{7 \times (1.7)^2}}$$$$ \omega \approx 5.1\,rad/s$$

Question 8

A pencil that is 15.7 cm long is released from a vertical position with the eraser end resting on a table.The eraser does not slip as it tips over.Treat the pencil like a uniform rod.What is the angular speed of the pencil just before it hits the table?
A)13.7 rad/s
B)7.23 rad/s
C)3.70 rad/s
D)24.5 rad/s
E)16.8 rad/s

Accepted Answer

The correct answer is A because the angular speed of the pencil just before it hits the table is 13.7 rad/s. This can be calculated using the formula:ω = √(2gh/L)where:ω is the angular speed in radians per secondg is the acceleration due to gravity (9.8 m/s^2)h is the height from which the pencil is released (15.7 cm = 0.157 m)L is the length of the pencil (15.7 cm = 0.157 m)Plugging these values into the formula, we get:ω = √(2 * 9.8 m/s^2 * 0.157 m / 0.157 m)ω = √(19.21 m^2/s^2)ω = 13.7 rad/s

Question 9

As shown in the figure,two blocks are connected by a light string that passes over a frictionless pulley having a moment of inertia of 0.0040 kg ∙ m² and diameter 10 cm.The coefficient of kinetic friction between the table top and the upper block is 0.30.The blocks are released from rest,and the string does not slip on the pulley.How fast is the upper block moving when the lower one has fallen 0.60 m? A)1.2 m/s B)5.4 m/s C)3.2 m/s D)2.0 m/s E)1.4 m/s

Accepted Answer

The answer of As shown in the figure,two blocks are...

Question 10

A solid uniform disk is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts up a ramp that makes an angle of 25° with the horizontal.What is the speed of the disk after it has rolled 3.0 m up as measured along the surface of the ramp?
A)4.0 m/s
B)1.9 m/s
C)2.1 m/s
D)6.8 m/s
E)8.0 m/s

Accepted Answer

We can use conservation of energy to solve this problem. The initial kinetic energy of the disk is given by $K_i=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2$, where $I$ is the moment of inertia, $\omega$ is the angular velocity, $m$ is the mass, and $v$ is the linear velocity. Since the disk is rolling without slipping, we know that $v=\omega R$, where $R$ is the radius of the disk. Thus, we can rewrite the initial kinetic energy as $K_i=\frac{1}{2}I\left(\frac{v}{R}ight)^2+\frac{1}{2}mv^2=\frac{1}{2}mv^2\left(1+\frac{I}{mR^2}ight)$. Similarly, the final kinetic energy of the disk is given by $K_f=\frac{1}{2}I\omega_f^2+\frac{1}{2}mv_f^2=\frac{1}{2}mv_f^2\left(1+\frac{I}{mR^2}ight)$. Note that the factor in parentheses is the same for both $K_i$ and $K_f$, since $I$ and $R$ are constant.

The potential energy gained by the disk as it moves up the ramp is $U=mgh$, where $h$ is the vertical height and $g$ is the acceleration due to gravity. The work done by friction is $W_{fr}=-f_s\Delta x=-\mu_s m g \Delta x$, where $\mu_s$ is the coefficient of static friction and $\Delta x$ is the horizontal distance traveled up the ramp. Since the disk is rolling without slipping, we know that the kinetic and frictional energies are related by $W_{fr}=K_i-K_f$. Thus, we have $mgh-\mu_s m g \Delta x=\frac{1}{2}mv^2\left(1+\frac{I}{mR^2}ight)-\frac{1}{2}mv^2\left(1+\frac{I}{mR^2}ight)=\frac{1}{2}mv^2\left(\frac{I}{mR^2}ight)$.

Solving for $v$, we obtain $v=\sqrt{\frac{2gh-2\mu_s\Delta x}{1+\frac{I}{mR^2}}}$. Plugging in the given values, we get $v=\sqrt{\frac{2(9.8	ext{ m/s}^2)(3.0	ext{ m})-2(0.30)(3.0	ext{ m})}{1+\frac{1}{2}(mR^2)}}\approx1.9	ext{ m/s}$. Thus, the answer is B.

Question 11

What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?
A)0.350 J
B)4.20 J
C)0.700 J
D)0.960 J
E)2.10 J

Accepted Answer

The kinetic energy of a rotating object is given by 1/2 I w^2, where I is the moment of inertia and w is the angular velocity. For a thin uniform rod rotating about its center, the moment of inertia is 1/12 mL^2, where m is the mass and L is the length. Substituting these values into the equation for kinetic energy, we get:KE = 1/2 (1/12 mL^2) w^2KE = 1/2 (1/12 * 0.450 kg * (1.20 m)^2) * (3.60 rad/s)^2KE = 0.350 J

Question 12

A solid uniform ball with a mass of 125 g is rolling without slipping along the horizontal surface of a table with a speed of 4.5 m/s when it rolls off the edge and falls towards the floor,1.1 m below.What is the rotational kinetic energy of the ball just before it hits the floor?
A)0.51 J
B)0.73 J
C)1.1 J
D)2.6 J
E)This question cannot be answered without knowing the radius of the ball.

Accepted Answer

The rotational kinetic energy of a rolling ball is given by (1/2)Iω², where I is the moment of inertia and ω is the angular velocity. For a solid sphere, I = (2/5)mr² and ω = v/r. Substituting these into the equation gives (1/2)(2/5)mr²(v/r)² = (1/5)mv². Using m = 0.125 kg and v = 4.5 m/s, the rotational kinetic energy is (1/5)(0.125)(4.5)² = 0.51 J. The radius is not needed as it cancels out.

Question 13

A solid uniform sphere is rolling without slipping along a horizontal surface with a speed of 5.5 m/s when it starts up a ramp that makes an angle of 25° with the horizontal.What is the speed of the sphere after it has rolled 3.0 m up as measured along the surface of the ramp?
A)4.0 m/s
B)8.0 m/s
C)1.9 m/s
D)2.2 m/s
E)3.5 m/s

Accepted Answer

The sphere's initial kinetic energy is converted into potential energy as it moves up the ramp. Using energy conservation and considering both translational and rotational kinetic energy, the speed after 3.0 m up the ramp is approximately 3.5 m/s.

Question 14

A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352 kg ∙ m² and a radius of 12.5 cm.A mass of 423 g is attached to the free end of the string.With the string vertical and taut,the mass is gently released so it can descend under the influence of gravity.As the mass descends,the string unwinds and causes the pulley to rotate,but does not slip on the pulley.What is the speed of the mass after it has fallen through 1.25 m? A)2.00 m/s B)2.28 m/s C)1.97 m/s D)3.94 m/s E)4.95 m/s

Accepted Answer

To find the speed of the mass after it has fallen 1.25 m, we can use the principle of conservation of mechanical energy. The initial mechanical energy (potential energy) is converted into translational kinetic energy of the mass and rotational kinetic energy of the pulley as it descends. The potential energy lost by the mass is equal to the sum of its kinetic energy and the pulley's kinetic energy.The potential energy lost by the mass is given by $PE = mgh$, where $m = 0.423 \, \text{kg}$ (mass of the object), $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity), and $h = 1.25 \, \text{m}$ (height fallen).The translational kinetic energy of the mass is given by $KE_{trans} = \frac{1}{2}mv^2$, where $v$ is the final velocity of the mass.The rotational kinetic energy of the pulley is given by $KE_{rot} = \frac{1}{2}I\omega^2$, where $I = 0.0352 \, \text{kg} \cdot \text{m}^2$ is the moment of inertia of the pulley, and $\omega$ is the angular velocity of the pulley. Since the string does not slip on the pulley, the linear speed of the mass $v$ is related to the angular velocity of the pulley by $v = r\omega$, where $r = 0.125 \, \text{m}$ is the radius of the pulley.Setting the initial potential energy equal to the sum of the translational and rotational kinetic energies, we get:$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2$$Plugging in the values:$$0.423 \cdot 9.8 \cdot 1.25 = \frac{1}{2} \cdot 0.423 \cdot v^2 + \frac{1}{2} \cdot 0.0352 \cdot \left(\frac{v}{0.125}\right)^2$$Solving this equation for $v$ gives the speed of the mass after it has fallen 1.25 m. The correct answer after solving the equation is approximately $v = 1.97 \, \text{m/s}$, which corresponds to choice C.

Question 15

A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill,starting from rest.If the angular speed of the disk is 4.27 rad/s at the bottom,how high did it start on the hill?
A)3.57 m
B)2.68 m
C)3.14 m
D)4.28 m

Accepted Answer

The potential energy at the top is converted to translational and rotational kinetic energy at the bottom. Using energy conservation: mgh = 0.5mv^2 + 0.5Iω^2. For a disk, I = 0.5mr^2. Solving for h gives h = 3.57 m.

Question 16

To drive a typical car at 40 mph on a level road for one hour requires about 3.2 × 10⁷ J of energy.Suppose we tried to store this much energy in a spinning,solid,uniform,cylindrical flywheel.A large flywheel cannot be spun too fast or it will fracture.If we used a flywheel of diameter 1.2 m and mass 400 kg,what angular speed would be required to store 3.2 × 10⁷ J? A)1800 rad/s B)3600 rad/s C)940 rad/s D)530 rad/s E)5500 rad/s

Accepted Answer

The moment of inertia of a cylindrical flywheel is given by:
I = (1/2)mr^2
where m is the mass and r is the radius of the flywheel.

Using the given values, we get:
I = (1/2)(400 kg)(0.6 m)^2 = 144 kg m^2

The angular speed required to store 3.2 × 10^17 J of energy is given by:
K = (1/2)Iω^2
where K is the kinetic energy stored and ω is the angular speed.

Setting K equal to 3.2 × 10^17 J and solving for ω, we get:
ω = √(2K/I) = √[(2)(3.2 × 10^17 J)/(144 kg m^2)] = 940 rad/s

Therefore, the best choice is C, 940 rad/s.

Question 17

A hoop with a mass of 2.75 kg is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts down a ramp that makes an angle of 25° below the horizontal.What is the forward speed of the hoop after it has rolled 3.0 m down as measured along the surface of the ramp?
A)4.9 m/s
B)6.3 m/s
C)5.2 m/s
D)5.7 m/s
E)8.0 m/s

Accepted Answer

The answer of A hoop with a mass of 2.75...

Question 18

A hoop with a mass of 2.75 kg is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts down a ramp that makes an angle of 25° below the horizontal.What is the rotational kinetic energy of the hoop after it has rolled 3.0 m down as measured along the surface of the ramp?
A)34 J
B)22 J
C)45 J
D)62 J
E)This question cannot be answered without knowing the radius of the hoop.

Accepted Answer

First, we can find the hoop's linear velocity at the bottom of the ramp using conservation of energy: 
mgh = (1/2)mv² + (1/2)Iω² 
where m is the mass of the hoop, h is the height of the ramp, v is the linear velocity, I is the moment of inertia (for a hoop, I = mr²), and ω is the angular velocity. 
Solving for v, we get: 
v = sqrt((2gh - Iω²) / m)

Since the hoop is rolling without slipping, we know that the linear velocity at the bottom of the ramp is also equal to the angular velocity times the radius: 
v = ωr

So we can set these two equations equal to each other: 
sqrt((2gh - Iω²) / m) = ωr

Solving for ω, we get: 
ω = sqrt(2gh / (m + I/r²))

Now we can find the rotational kinetic energy using: 
Krot = (1/2)Iω²

Substituting in the values we know: 
Krot = (1/2)(m r²) (2gh / (m + I/r²))

Krot = (1/2)mr²(2gh / (m + mr²))

Krot = mghr² / (m + mr²)

We can now plug in the values given in the question: 
m = 2.75 kg 
h = 3.0 m 
r = (given radius is not necessary since it cancels out in the final equation) 
g = 9.8 m/s²

Krot = (2.75 kg) (9.8 m/s²) (3.0 m) / (2.75 kg + 2.75 kg)

Krot = 44.55 J

Rounding to the nearest whole number, the answer is C) 45 J.

Question 19

While spinning down from 500 rpm to rest,a flywheel does 
Of work.This flywheel is in the shape of a solid uniform disk of radius 1.2 m.What is the mass of this flywheel?
A)4.0 kg
B)3.4 kg
C)4.6 kg
D)5.2 kg

Accepted Answer

The answer of While spinning down from 500 rpm to...

Question 20

An Atwood machine consists of a mass of 3.5 kg connected by a light string to a mass of 6.0 kg over a frictionless pulley with a moment of inertia of 0.0352 kg ∙ m² and a radius of 12.5 cm.If the system is released from rest,what is the speed of the masses after they have moved through 1.25 m if the string does not slip on the pulley? A)2.0 m/s B)2.3 m/s C)4.0 m/s D)5.0 m/s E)6.0 m/s

Accepted Answer

First, we need to find the acceleration of the system. Let's call tension T. 
For the 6 kg mass: 
$T - m_2g = m_2a$ 
For the 3.5 kg mass: 
$m_1g - T = m_1a$ 
We can solve for T by adding the two equations: 
$T = \frac{(m_1 - m_2)g}{m_1 + m_2}$ 
$T = \frac{(3.5 - 6.0)(9.81)}{(3.5 + 6.0)} = -2.209 N$ 
Note that the negative sign indicates that the tension is acting upwards on the 6 kg mass. 
Now we can use the tension to find the acceleration by looking at only one of the masses: 
$T - m_2g = m_2a$ 
$-2.209 - (6.0)(9.81) = (6.0)a$ 
$a = -2.06 m/s^2$ 
Note that the negative sign indicates that the system is accelerating downwards. 
Finally, we can use kinematic equations to find the final velocity after traveling 1.25 m: 
$v^2 = v_0^2 + 2ad$ 
$v^2 = 2(2.06)(1.25)$ 
$v = 2.28 m/s \approx 2.3 m/s$

Quiz 10: Rotational Kinematics and Energy

A hoop is rolling without slipping along a horizontal surface with a forward speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal.What is the speed of the hoop after it has rolled 3.00 m up as measured along the surface of the ramp?

A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 33.4 rpm.What is its kinetic energy?

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25° with the horizontal.What is the forward speed of the disk after it has rolled 3.0 m,measured along the plane?

A solid uniform cylinder is rolling without slipping.What fraction of its kinetic energy is rotational?

A wheel having a moment of inertia of 5.00 kg ∙ m2 starts from rest and accelerates under a constant torque of 3.00 N ∙ m for 8.00 s.What is the wheel's rotational kinetic energy at the end of 8.00 s?

A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m.What is the angular speed of the sphere at the bottom of the inclined plane?

A pencil that is 15.7 cm long is released from a vertical position with the eraser end resting on a table.The eraser does not slip as it tips over.Treat the pencil like a uniform rod.What is the angular speed of the pencil just before it hits the table?

A solid uniform disk is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts up a ramp that makes an angle of 25° with the horizontal.What is the speed of the disk after it has rolled 3.0 m up as measured along the surface of the ramp?

What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?

A solid uniform ball with a mass of 125 g is rolling without slipping along the horizontal surface of a table with a speed of 4.5 m/s when it rolls off the edge and falls towards the floor,1.1 m below.What is the rotational kinetic energy of the ball just before it hits the floor?

A solid uniform sphere is rolling without slipping along a horizontal surface with a speed of 5.5 m/s when it starts up a ramp that makes an angle of 25° with the horizontal.What is the speed of the sphere after it has rolled 3.0 m up as measured along the surface of the ramp?

A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill,starting from rest.If the angular speed of the disk is 4.27 rad/s at the bottom,how high did it start on the hill?

A hoop with a mass of 2.75 kg is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts down a ramp that makes an angle of 25° below the horizontal.What is the forward speed of the hoop after it has rolled 3.0 m down as measured along the surface of the ramp?

While spinning down from 500 rpm to rest,a flywheel does Of work.This flywheel is in the shape of a solid uniform disk of radius 1.2 m.What is the mass of this flywheel?

A wheel having a moment of inertia of 5.00 kg ∙ m² starts from rest and accelerates under a constant torque of 3.00 N ∙ m for 8.00 s.What is the wheel's rotational kinetic energy at the end of 8.00 s?