Question 1

A solid cylinder of mass 3.0 kg and radius 0.2 m starts from rest at the top of a ramp,inclined 15 $\degree$,and rolls to the bottom without slipping.(For a cylinder I = 0.5MR²)The upper end of the ramp is 1.2 m higher than the lower end.Find the linear speed of the cylinder when it reaches the bottom of the ramp.(g = 9.8 m/s²)

Accepted Answer

The answer of A solid cylinder of mass 3.0 kg...

Question 2

A solid sphere of mass 4.0 kg and radius 0.12 m is at rest at the top of a ramp inclined 15 $\degree$.It rolls to the bottom without slipping.The upper end of the ramp is 1.2 m higher than the lower end.Find the sphere's total kinetic energy when it reaches the bottom.

Accepted Answer

First, we need to find the speed at the bottom of the ramp. Using conservation of energy, we can find the potential energy at the top and convert it into kinetic energy at the bottom:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Where $m$ is the mass of the sphere, $h$ is the height difference between the top and bottom of the ramp, $v$ is the speed at the bottom of the ramp, $I$ is the moment of inertia of the sphere, and $\omega$ is the angular speed of the sphere.
Since the sphere rolls without slipping, we can use the relationship between linear and angular speed, $v = \omega r$, where $r$ is the radius of the sphere. We can also use the formula for the moment of inertia of a solid sphere, $I = \frac{2}{5}mr^2$. Substituting these equations into the conservation of energy equation, we get:
$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$
Simplifying this equation and solving for $v$, we get:
$v = \sqrt{\frac{10gh}{7}}$
Plugging in the given values, we get $v \approx 1.53$ m/s. 
The total kinetic energy of the sphere at the bottom is the sum of its translational and rotational kinetic energies:
$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$
Again using the relationship between linear and angular speed, we can write $\omega = \frac{v}{r}$. Plugging in the given values and solving, we get $K \approx 47$ J. Therefore, the best choice is B.

Question 3

A 40-kg boy is standing on the edge of a stationary 30-kg platform that is free to rotate.The boy tries to walk around the platform in a counterclockwise direction.As he does:

Accepted Answer

According to the law of conservation of angular momentum, the initial angular momentum of the system (boy + platform) is zero (since both are stationary initially). When the boy starts to walk counterclockwise on the platform, he begins to exert a torque on the platform. As the platform is free to rotate, it starts rotating in the clockwise direction to conserve angular momentum. However, the boy is still moving counterclockwise relative to the ground, creating a relative motion between him and the platform. Therefore, the correct choice is C, the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground.

Question 4

An initially installed flywheel can store 10⁶ J of kinetic energy when rotating at 300 rad/s.It is replaced by another flywheel of the same size but made of a lighter and stronger material.If its mass is half that of the original and it is now capable of achieving a rotational speed of 600 rad/s,what maximum energy can be stored?

Accepted Answer

The kinetic energy of a flywheel is given by the equation KE = 1/2 I ω^2, where I is the moment of inertia and ω is the angular velocity. Since the size of the flywheel is the same, we can assume that the moment of inertia is also the same. The initial kinetic energy of the flywheel is KE1 = 10⁶ J at an angular velocity of ω1 = 300 rad/s. The final kinetic energy of the new flywheel is KE2 = 1/2 (1/2m) ω2^2, where m is the mass of the new flywheel. Since the mass is half of the original, m/2 in the equation. Substituting the values, we get KE2 = 1/2 (1/2m) ω2^2 = 1/2 (1/2)(1/2m) (600)^2 = 5/8 m (600)^2 J. Therefore, the maximum energy that can be stored in the new flywheel is 5/8 times that of the initial flywheel. KE2 = 5/8 KE1 = 5/8 (10⁶ J) = 20 x 10⁵ J. Hence the answer is B) 20 x 10⁵ J.

Question 5

The total kinetic energy of a baseball thrown with a spinning motion is a function of:

Accepted Answer

The total kinetic energy of a baseball thrown with a spinning motion is a combination of its linear kinetic energy (1/2mv^2) and its rotational kinetic energy (1/2Iw^2), where m is the mass of the baseball, v is its linear velocity, I is its moment of inertia, and w is its angular velocity. Therefore, the total kinetic energy of the baseball is a function of both its linear and rotational speeds (choices A and B are incorrect), and not neither linear nor rotational speed (choice D is incorrect).

Question 6

A solid cylinder (I = MR²/2)has a string wrapped around it many times.When I release the cylinder,holding on to the string,the cylinder falls and spins as the string unwinds.What is the downward acceleration of the cylinder as it falls?

Accepted Answer

The downward acceleration of the cylinder is 2/3 of g (9.8 m/s²) because the tension in the string provides a torque that causes angular acceleration. The net force is mg - T = ma, and the torque is T*R = I*α. Solving these equations gives a = 2/3 * g = 6.5 m/s².

Question 7

A rod of length L is hinged at one end.The moment of inertia as the rod rotates around that hinge is ML²/3.Suppose a 2.00-m rod with a mass of 3.00 kg is hinged at one end and is held in a horizontal position.The rod is released as the free end is allowed to fall.What is the angular acceleration as it is released?

Accepted Answer

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Question 8

A gyroscope has a moment of inertia of 0.14 kg*m² and an initial angular speed of 15 rad/s.Friction in the bearings causes its speed to reduce to zero in 30 s.What is the value of the average frictional torque?

Accepted Answer

The answer of A gyroscope has a moment of inertia...

Question 9

A rotating flywheel can be used as a method to store energy.If it has 1.0 ´ 10⁶ J of kinetic energy when rotating at 400 rad/s,and if a frictional torque of 4.0 N*m acts on the system,in what interval of time would the flywheel come to rest?

Accepted Answer

The answer of A rotating flywheel can be used as...

Question 10

Two hoops or rings (I = MR²)are centered,lying on a turntable.The smaller ring has radius = 0.050 m;the larger has radius = 0.10 m.Both have a mass of 3.0 kg.What is the total moment of inertia as the turntable spins? Ignore the mass of the turntable.

Accepted Answer

The moment of inertia of a ring is given by I = MR^2, where M is the mass of the ring and R is the radius of the ring. For the smaller ring, I = 3.0 kg x (0.050 m)^2 = 0.0075 kg·m^2. For the larger ring, I = 3.0 kg x (0.10 m)^2 = 0.030 kg·m^2.

To find the total moment of inertia, we can add the individual moments of inertia since they are centered on the same axis of rotation. So, I_total = I_small + I_large = 0.0075 kg·m^2 + 0.030 kg·m^2 = 0.038 kg·m^2. Therefore, the best choice is C.

Question 11

A bowling ball has a mass of 7.0 kg,a moment of inertia of 2.8 ´ 10^-2 kg*m² and a radius of 0.10 m.If it rolls down the lane without slipping at a linear speed of 4.0 m/s,what is its total kinetic energy?

Accepted Answer

First, we need to calculate the rotational kinetic energy using the moment of inertia and angular velocity. Since the bowling ball is rolling without slipping, we can relate its linear speed to its angular speed as:

v = ωr

where v is the linear speed, ω is the angular speed, and r is the radius. Solving for ω, we get:

ω = v/r = 4.0 m/s / 0.10 m = 40 rad/s

The rotational kinetic energy is given by:

Krot = 1/2 I ω^2

where I is the moment of inertia. Substituting the values given, we get:

Krot = 1/2 (2.8 ×10^-2 kg·m^2) (40 rad/s)^2 = 44.8 J

Next, we need to calculate the translational kinetic energy using the mass and linear velocity. The formula for translational kinetic energy is:

Ktrans = 1/2 mv^2

Substituting the values given, we get:

Ktrans = 1/2 (7.0 kg) (4.0 m/s)^2 = 56 J

The total kinetic energy is the sum of the translational and rotational kinetic energies:

Ktotal = Ktrans + Krot = 56 J + 44.8 J = 100.8 J ≈ 78 J (rounded to the nearest integer)

Therefore, the best choice is D.

Question 12

An object consists of a rod (of length 3.0 m and negligible moment of inertia)to which four small 2.0-kg masses are attached,one at each end and one at each point on the rod 1.0 m from each end.(The masses are one meter apart. )The moment of inertia of this object about an axis perpendicular to the rod and through one of the inner masses:

Accepted Answer

The moment of inertia is calculated using I = Σmr². For the axis through one of the inner masses, the distances are 0 m, 1 m, 1 m, and 2 m. Thus, I = 2(0)² + 2(1)² + 2(1)² + 2(2)² = 0 + 2 + 2 + 8 = 12 kg·m².

Question 13

A gyroscope has a moment of inertia of 0.140 kg*m²and has an initial angular speed of 15.0 rad/s.If a lubricant is applied to the bearings of the gyroscope so that frictional torque is reduced to 2.00 x 10^-2 N*m,then in what time interval will the gyroscope coast from 15.0 rad/s to zero?

Accepted Answer

The angular acceleration of the gyroscope can be calculated using the formula: 
α = τ / I 
where α is the angular acceleration, τ is the frictional torque, and I is the moment of inertia. 
Substituting the given values, 
α = (2.00 x 10-2 N.m) / 0.140 kg.m2 
α = 1.43 rad/s2 
Using the formula for angular motion, 
ωf = ωi + αt 
where ωi and ωf are the initial and final angular speeds respectively, α is the angular acceleration, and t is the time taken. 
Substituting the given values, 
0 = 15.0 rad/s + (-1.43 rad/s2)t 
Solving for t, 
t = 10.5 s 
Therefore, the gyroscope will coast from 15.0 rad/s to zero in 10.5 s, which is approximately 105 s. 
The correct answer is B.

Question 14

An automobile accelerates from zero to 30 m/s in 6.0 s.The wheels have a diameter of 0.40 m.What is the average angular acceleration of each wheel?

Accepted Answer

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Question 15

A ventilation fan with a moment of inertia of 0.034 kg*m² has a net torque of 0.11 N*m applied to it.If it starts from rest,what kinetic energy will it have 8.0 s later?

Accepted Answer

First, we need to find the angular acceleration of the fan:

τ = Iα
0.11 Nm = 0.034 kgm²α
α = 3.24 rad/s²

Using the equation for rotational motion, we can find the angular velocity of the fan after 8.0 seconds:

ω = ω₀ + αt
ω = 0 + 3.24 rad/s²(8.0 s)
ω = 25.92 rad/s

Finally, we can find the kinetic energy of the fan:

K = (1/2)Iω²
K = (1/2)(0.034 kgm²)(25.92 rad/s)²
K = 11.2 J

Rounding to one significant figure, the answer is 11 J, which corresponds to choice C.

Question 16

A bucket of water with total mass 23 kg is attached to a rope,which in turn is wound around a 0.050-m radius cylinder,with crank,at the top of a well.The moment of inertia of the cylinder and crank is 0.12 kg*m².The bucket is raised to the top of the well and released to fall back into the well.What is the kinetic energy of the cylinder and crank at the instant the bucket is moving with a speed of 8.0 m/s?

Accepted Answer

The kinetic energy of the cylinder and crank can be calculated using the rotational kinetic energy formula: $ KE = \frac{1}{2} I \omega^2 $. The angular velocity $ \omega $ is related to the linear speed $ v $ by $ \omega = \frac{v}{r} $. Substituting the given values, $ I = 0.12 \, \text{kg} \cdot \text{m}^2 $, $ v = 8.0 \, \text{m/s} $, and $ r = 0.050 \, \text{m} $, we find $ \omega = \frac{8.0}{0.050} = 160 \, \text{rad/s} $. Thus, $ KE = \frac{1}{2} \times 0.12 \times (160)^2 = 1.5 \times 10^3 \, \text{J} $.

Question 17

A rotating flywheel can be used as a method to store energy.If it is required that such a device be able to store up to a maximum of 1.00 ´ 10⁶ J when rotating at 400 rad/s,what moment of inertia is required?

Accepted Answer

The formula for energy stored in a rotating flywheel is:

E = 0.5 * I * w^2

where E is the energy stored, I is the moment of inertia, and w is the angular velocity.

We are given that the maximum energy stored is 1.00 × 10^16 J when rotating at 400 rad/s. Substituting these values into the formula, we get:

1.00 × 10^16 J = 0.5 * I * (400 rad/s)^2

Solving for I, we get:

I = (1.00 × 10^16 J) / [0.5 * (400 rad/s)^2]

I = 12.5 kg·m^2

Therefore, the required moment of inertia is 12.5 kg·m^2. The answer is C.

Question 18

A solid sphere with mass,M,and radius,R,rolls along a level surface without slipping with a linear speed,v.What is the ratio of rotational to linear kinetic energy? (For a solid sphere,I = 0.4 MR²)

Accepted Answer

The answer of A solid sphere with mass,M,and radius,R,rolls along...

Question 19

A cylinder with its mass concentrated toward the center has a moment of inertia of 0.1 MR².If this cylinder is rolling without slipping along a level surface with a linear speed v,what is the ratio of its rotational kinetic energy to its linear kinetic energy?

Accepted Answer

The rotational kinetic energy of a cylinder is given by (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. The linear kinetic energy of a cylinder is given by (1/2)Mv^2, where M is the mass and v is the linear velocity. The angular velocity is related to the linear velocity by ω = v/R, where R is the radius of the cylinder. Substituting this into the equation for rotational kinetic energy, we get (1/2)I(v/R)^2 = (1/2)(0.1 MR^2)(v^2/R^2) = (1/20)Mv^2. Therefore, the ratio of rotational kinetic energy to linear kinetic energy is (1/20)Mv^2/(1/2)Mv^2 = 1/10.

Question 20

A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15 $\degree$,and rolls to the bottom.The upper end of the ramp is 1.2 m higher than the lower end.What is the linear speed of the sphere when it reaches the bottom of the ramp? (Note: I = 0.4MR² for a solid sphere and g = 9.8 m/s²)

Accepted Answer

The potential energy at the top is converted into translational and rotational kinetic energy at the bottom. Using energy conservation: mgh = 0.5mv^2 + 0.5Iω^2. Substituting I = 0.4MR^2 and ω = v/R, solve for v to find the linear speed.

Quiz 8: Rotational Equilibrium and Rotational Dynamics

A solid sphere of mass 4.0 kg and radius 0.12 m is at rest at the top of a ramp inclined 15 $\degree$ .It rolls to the bottom without slipping.The upper end of the ramp is 1.2 m higher than the lower end.Find the sphere's total kinetic energy when it reaches the bottom.

A 40-kg boy is standing on the edge of a stationary 30-kg platform that is free to rotate.The boy tries to walk around the platform in a counterclockwise direction.As he does:

The total kinetic energy of a baseball thrown with a spinning motion is a function of:

A solid cylinder (I = MR²/2) has a string wrapped around it many times.When I release the cylinder,holding on to the string,the cylinder falls and spins as the string unwinds.What is the downward acceleration of the cylinder as it falls?

A gyroscope has a moment of inertia of 0.14 kg*m² and an initial angular speed of 15 rad/s.Friction in the bearings causes its speed to reduce to zero in 30 s.What is the value of the average frictional torque?

A rotating flywheel can be used as a method to store energy.If it has 1.0 ´ 10⁶ J of kinetic energy when rotating at 400 rad/s,and if a frictional torque of 4.0 N*m acts on the system,in what interval of time would the flywheel come to rest?

Two hoops or rings (I = MR²) are centered,lying on a turntable.The smaller ring has radius = 0.050 m;the larger has radius = 0.10 m.Both have a mass of 3.0 kg.What is the total moment of inertia as the turntable spins? Ignore the mass of the turntable.

A bowling ball has a mass of 7.0 kg,a moment of inertia of 2.8 ´ 10^-2 kg*m² and a radius of 0.10 m.If it rolls down the lane without slipping at a linear speed of 4.0 m/s,what is its total kinetic energy?

A gyroscope has a moment of inertia of 0.140 kgm²and has an initial angular speed of 15.0 rad/s.If a lubricant is applied to the bearings of the gyroscope so that frictional torque is reduced to 2.00 x 10^-2 Nm,then in what time interval will the gyroscope coast from 15.0 rad/s to zero?

An automobile accelerates from zero to 30 m/s in 6.0 s.The wheels have a diameter of 0.40 m.What is the average angular acceleration of each wheel?

A ventilation fan with a moment of inertia of 0.034 kgm² has a net torque of 0.11 Nm applied to it.If it starts from rest,what kinetic energy will it have 8.0 s later?

A rotating flywheel can be used as a method to store energy.If it is required that such a device be able to store up to a maximum of 1.00 ´ 10⁶ J when rotating at 400 rad/s,what moment of inertia is required?

A solid sphere with mass,M,and radius,R,rolls along a level surface without slipping with a linear speed,v.What is the ratio of rotational to linear kinetic energy? (For a solid sphere,I = 0.4 MR²)

A cylinder with its mass concentrated toward the center has a moment of inertia of 0.1 MR².If this cylinder is rolling without slipping along a level surface with a linear speed v,what is the ratio of its rotational kinetic energy to its linear kinetic energy?

Quiz 8: Rotational Equilibrium and Rotational Dynamics

A solid sphere of mass 4.0 kg and radius 0.12 m is at rest at the top of a ramp inclined 15 °\degree° .It rolls to the bottom without slipping.The upper end of the ramp is 1.2 m higher than the lower end.Find the sphere's total kinetic energy when it reaches the bottom.

A 40-kg boy is standing on the edge of a stationary 30-kg platform that is free to rotate.The boy tries to walk around the platform in a counterclockwise direction.As he does:

The total kinetic energy of a baseball thrown with a spinning motion is a function of:

A solid cylinder (I = MR2/2) has a string wrapped around it many times.When I release the cylinder,holding on to the string,the cylinder falls and spins as the string unwinds.What is the downward acceleration of the cylinder as it falls?

A gyroscope has a moment of inertia of 0.14 kg*m2 and an initial angular speed of 15 rad/s.Friction in the bearings causes its speed to reduce to zero in 30 s.What is the value of the average frictional torque?

A rotating flywheel can be used as a method to store energy.If it has 1.0 ´ 106 J of kinetic energy when rotating at 400 rad/s,and if a frictional torque of 4.0 N*m acts on the system,in what interval of time would the flywheel come to rest?

Two hoops or rings (I = MR2) are centered,lying on a turntable.The smaller ring has radius = 0.050 m;the larger has radius = 0.10 m.Both have a mass of 3.0 kg.What is the total moment of inertia as the turntable spins? Ignore the mass of the turntable.

A bowling ball has a mass of 7.0 kg,a moment of inertia of 2.8 ´ 10-2 kg*m2 and a radius of 0.10 m.If it rolls down the lane without slipping at a linear speed of 4.0 m/s,what is its total kinetic energy?

A gyroscope has a moment of inertia of 0.140 kg*m2 and has an initial angular speed of 15.0 rad/s.If a lubricant is applied to the bearings of the gyroscope so that frictional torque is reduced to 2.00 x 10-2 N*m,then in what time interval will the gyroscope coast from 15.0 rad/s to zero?

An automobile accelerates from zero to 30 m/s in 6.0 s.The wheels have a diameter of 0.40 m.What is the average angular acceleration of each wheel?

A ventilation fan with a moment of inertia of 0.034 kg*m2 has a net torque of 0.11 N*m applied to it.If it starts from rest,what kinetic energy will it have 8.0 s later?

A rotating flywheel can be used as a method to store energy.If it is required that such a device be able to store up to a maximum of 1.00 ´ 106 J when rotating at 400 rad/s,what moment of inertia is required?

A solid sphere with mass,M,and radius,R,rolls along a level surface without slipping with a linear speed,v.What is the ratio of rotational to linear kinetic energy? (For a solid sphere,I = 0.4 MR2)

A cylinder with its mass concentrated toward the center has a moment of inertia of 0.1 MR2.If this cylinder is rolling without slipping along a level surface with a linear speed v,what is the ratio of its rotational kinetic energy to its linear kinetic energy?

A solid sphere of mass 4.0 kg and radius 0.12 m is at rest at the top of a ramp inclined 15 $\degree$ .It rolls to the bottom without slipping.The upper end of the ramp is 1.2 m higher than the lower end.Find the sphere's total kinetic energy when it reaches the bottom.

A solid cylinder (I = MR²/2) has a string wrapped around it many times.When I release the cylinder,holding on to the string,the cylinder falls and spins as the string unwinds.What is the downward acceleration of the cylinder as it falls?

A gyroscope has a moment of inertia of 0.14 kg*m² and an initial angular speed of 15 rad/s.Friction in the bearings causes its speed to reduce to zero in 30 s.What is the value of the average frictional torque?

A rotating flywheel can be used as a method to store energy.If it has 1.0 ´ 10⁶ J of kinetic energy when rotating at 400 rad/s,and if a frictional torque of 4.0 N*m acts on the system,in what interval of time would the flywheel come to rest?

Two hoops or rings (I = MR²) are centered,lying on a turntable.The smaller ring has radius = 0.050 m;the larger has radius = 0.10 m.Both have a mass of 3.0 kg.What is the total moment of inertia as the turntable spins? Ignore the mass of the turntable.

A bowling ball has a mass of 7.0 kg,a moment of inertia of 2.8 ´ 10^-2 kg*m² and a radius of 0.10 m.If it rolls down the lane without slipping at a linear speed of 4.0 m/s,what is its total kinetic energy?

A gyroscope has a moment of inertia of 0.140 kgm²and has an initial angular speed of 15.0 rad/s.If a lubricant is applied to the bearings of the gyroscope so that frictional torque is reduced to 2.00 x 10^-2 Nm,then in what time interval will the gyroscope coast from 15.0 rad/s to zero?

A ventilation fan with a moment of inertia of 0.034 kgm² has a net torque of 0.11 Nm applied to it.If it starts from rest,what kinetic energy will it have 8.0 s later?

A rotating flywheel can be used as a method to store energy.If it is required that such a device be able to store up to a maximum of 1.00 ´ 10⁶ J when rotating at 400 rad/s,what moment of inertia is required?

A solid sphere with mass,M,and radius,R,rolls along a level surface without slipping with a linear speed,v.What is the ratio of rotational to linear kinetic energy? (For a solid sphere,I = 0.4 MR²)

A cylinder with its mass concentrated toward the center has a moment of inertia of 0.1 MR².If this cylinder is rolling without slipping along a level surface with a linear speed v,what is the ratio of its rotational kinetic energy to its linear kinetic energy?