Question 1

A homogeneous solid cylinder of mass m,length L,and radius R rotates about an axis through point P,which is parallel to the cylinder axis.If the moment of inertia about the cylinder axis is mR²,the moment of inertia about the axis through P is

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Question 2

The disc brakes of a high performance car are often made of carbon fiber instead of iron,thereby reducing the mass.If both types of discs are of the same size and shape,and each iron disc has a mass of 4 kg and each carbon disc has a mass of 1 kg,what is the reduction in rotational kinetic energy at 72 km/h if all the four iron discs in the car are replaced with carbon discs?

Accepted Answer

The reduction in mass for each disc is 3 kg (4 kg - 1 kg). For four discs, the total mass reduction is 12 kg. The reduction in rotational kinetic energy is calculated using the formula: ΔKE = 0.5 * ΔI * ω², where ΔI is the change in moment of inertia and ω is the angular velocity. Since the discs are the same size, ΔI = Δm * r². Assuming the car's wheels have a radius r, the reduction in rotational kinetic energy is 0.5 * 12 kg * (v/r)², where v = 72 km/h = 20 m/s. Simplifying, the reduction is 1200 J.

Question 3

In the figure,R₁ = R₂ and cm is the center of mass.The rotational inertia about an axis through point P₁ is I₁,the rotational inertia about an axis through point P₂ is I₂,and the rotational inertia about an axis through the cm is I_cm.The relationship among the moments is

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Question 4

A body of mass m is whirled at a constant angular velocity on the end of a string of length R.To double the kinetic energy of the body as it whirls while maintaining the angular velocity,the length of the string must be changed to

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Question 5

A ball of mass m₁,connected to another mass m₂ by a string,is whirled at a constant speed in a horizontal circle of radius R equal to 0.800m.If the mass m₂ = 5.00 kg,the kinetic energy of the ball is

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Question 6

To increase the moment of inertia of a body about an axis,you must

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Question 7

The amount of work done on a rotating body can be expressed in terms of the product of

Accepted Answer

The amount of work done on a rotating body can be expressed in terms of the product of torque and angular displacement. This is because work is a measure of the energy transferred to an object and in rotation, torque is what transfers energy to the object, while angular displacement is the motion of the object around the axis of rotation. Therefore, the product of these two quantities gives us the amount of work done on the rotating body.

Question 8

The disc brake of a high performance car is often made of carbon fiber instead of iron,thereby reducing the mass.If the mass of a carbon disc is  that of iron for the same shape and size,by what factor is the moment of inertia changed?

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Question 9

The rotational inertia of an object about an axis depends on the

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The rotational inertia (also known as moment of inertia) of an object about an axis depends on the mass distribution of the object with respect to that axis. It is a measure of the resistance of an object to rotational motion around that axis. The other choices are not directly related to rotational inertia; angular velocity and acceleration are affected by the moment of inertia, torque is the product of moment of inertia and angular acceleration, and linear acceleration has no effect on rotational inertia.

Question 10

The moment of inertia of a slim rod about a transverse axis through one end is mL²/3,where m is the mass of the rod and L is its length.The moment of inertia of a 0.24-kg meterstick about a transverse axis through its center is

Accepted Answer

Using the given formula, I = mL²/3, and substituting in the given values for the mass and length of the meterstick, we get: I = (0.24 kg) × (1 m) × (0.5 m)2/3 = 0.020 kg · m² Therefore, the moment of inertia of the meterstick about a transverse axis through its center is 0.020 kg · m² (option C).

Question 11

A stone of mass 10 kg has a rotational inertia of 2.4 kg · m² about an axis A parallel to an axis through the center of mass.If axis A is 0.20m from the center of mass axis,the rotational inertia about the center of mass axis is

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Question 12

Four 50-g point masses are at the corners of a square with 20-cm sides.What is the moment of inertia of this system about an axis perpendicular to the plane of the square and passing through its center?

Accepted Answer

The moment of inertia for point masses is calculated as I = Σmr². Each mass is 50 g (0.05 kg) and the distance from the center to each corner is √(0.1² + 0.1²) = √0.02 m. Thus, I = 4 * 0.05 * 0.02 = 4.0 × 10⁻³ kg·m².

Question 13

A hoop of mass 50 kg rolls without slipping.If the center-of-mass of the hoop has a translational speed of 4.0 m/s,the total kinetic energy of the hoop is

Accepted Answer

The total kinetic energy of a rolling object (without slipping) is the sum of its translational kinetic energy and rotational kinetic energy. The translational kinetic energy ($K_{trans}$) is given by $\frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity of the center of mass. The rotational kinetic energy ($K_{rot}$) about the center of mass for a hoop is given by $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia of the hoop about its center ($I = mr^2$ for a hoop) and $\omega$ is the angular velocity. Since the hoop rolls without slipping, $v = r\omega$, where $r$ is the radius of the hoop.Given: $m = 50\,kg$, $v = 4.0\,m/s$1. Calculate $K_{trans} = \frac{1}{2}mv^2 = \frac{1}{2} \times 50 \times (4.0)^2 = 400\,J$2. To find $K_{rot}$, we use $v = r\omega$ to express $\omega$ as $\omega = \frac{v}{r}$. Then, $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times (mr^2) \times \left(\frac{v}{r}\right)^2 = \frac{1}{2} \times 50 \times 4^2 = 400\,J$Thus, the total kinetic energy $K_{total} = K_{trans} + K_{rot} = 400\,J + 400\,J = 800\,J$, which is not listed among the options provided.

Question 14

A 7.00-kg mass and a 4.00-kg mass are mounted on a spindle free to turn about the x axis as shown.Assume the mass of the arms and the spindle to be negligible.The rotational inertia of this system is approximately

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Question 15

The moment of inertia of a slim rod of mass m and length L about a transverse axis through one end is mL²/3.The moment of inertia of such a rod about a transverse axis through the rod at a distance L/3 from one end is

Accepted Answer

The moment of inertia of a slim rod of mass m and length L about a transverse axis through one end is given as: I = mL^2/3 We can use the parallel axis theorem to find the moment of inertia of the same rod about a transverse axis through the rod at a distance L/3 from one end: I' = I + md^2, where d is the distance between the two axes. d = L - L/3 = 2L/3 Therefore, I' = mL^2/3 + m(2L/3)^2 = mL^2/3 + 4mL^2/9 = 7mL^2/9 = mL²/9 Hence, the correct choice is (C).

Question 16

A spherical shell and a solid sphere (both of the same total mass)are rotating about a diameter at the same angular frequency and with the same rotational kinetic energy.What is the ratio of the spherical shell radius to the solid sphere (i.e.,R_{spherical shell}/R_{solid sphere})?

Accepted Answer

The rotational kinetic energy of a solid sphere is (2/5)MR^2ω^2, while that of a spherical shell is (1/2)MR^2ω^2. Since the two spheres have the same total mass and rotational kinetic energy, we have (2/5)MR^2ω^2 = (1/2)MR^2ω^2, which gives R_spherical shell/R_solid sphere = (3/5)^1/2.

Question 17

A solid sphere of radius r = 5 cm and mass m = 0.2 kg rolls in a groove as shown.The angle,$\theta$,between the points of contact with the groove is 120S1U1P1$\circ$S1S1P0.If the sphere has a linear velocity of 10 mm/s,what is the rotational kinetic energy?

Accepted Answer

The rotational kinetic energy of a sphere rolling without slipping is given by

KE_rotational = (2/5) * m * R^2 * w^2

where m is the mass of the sphere, R is the radius of the sphere, and w is the angular velocity of the sphere.

To find the angular velocity w, we can use the fact that the linear velocity of the sphere is related to its rotational motion by

v = R*w

where v is the linear velocity of the sphere. Plugging in the given values, we get

10 mm/s = 0.01 m/s = R*w

w = 0.01 m/s / R = 0.01 m/s / 0.05 m = 0.2 rad/s

Now we can plug in all the values to find the rotational kinetic energy:

KE_rotational = (2/5) * 0.2 kg * (0.05 m)^2 * (0.2 rad/s)^2 = 0.0016 J = 1.6 mJ

Therefore, the correct answer is choice C.

Question 18

A thin solid disk of radius R = 0.5 m and mass M = 2.0 kg is rolling without slipping on a horizontal surface with a linear speed v = 5.0 m/s.The disk now rolls without slipping up an inclined plane that is at an angle of 60 degrees to the vertical.Calculate the maximum height that the disk rolls up the incline.

Accepted Answer

The maximum height $h$ the disk rolls up the incline can be found using energy conservation. The initial kinetic energy (both translational and rotational) is converted into potential energy as the disk climbs the incline. The initial kinetic energy $K_{\text{initial}}$ is the sum of translational kinetic energy $\frac{1}{2}Mv^2$ and rotational kinetic energy $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia of the disk ($\frac{1}{2}MR^2$ for a solid disk) and $\omega$ is the angular velocity ($\frac{v}{R}$ for rolling without slipping). The potential energy at height $h$ is $Mgh$, where $g$ is the acceleration due to gravity (9.8 m/s$^2$). Setting the initial kinetic energy equal to the potential energy gives:$$\frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v}{R}\right)^2 = Mgh$$Simplifying and solving for $h$ gives:$$h = \frac{v^2}{2g}\left(1 + \frac{1}{2}\right) = \frac{5^2}{2 \times 9.8}\left(\frac{3}{2}\right) \approx 2.6 \text{ m}$$Therefore, the maximum height the disk rolls up the incline is approximately 2.6 meters.

Question 19

The moment of inertia of a set of dumbbells,considered as two mass points m separated by a distance 2L about the axis AA,is

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Question 20

A wagon wheel consists of 8 spokes of uniform diameter,each of mass m_s and length L cm.The outer ring has a mass m_ring.What is the moment of inertia of the wheel? Assume that each spoke extends from the center to the other ring and the ring is of negligible thickness.

Accepted Answer

The moment of inertia of the wagon wheel can be found by considering the moment of inertia of each spoke, as well as the moment of inertia of the outer ring.

The moment of inertia of each spoke is given by: 
Ispoke = (1/12) m L^2

Since there are 8 spokes, the total moment of inertia due to the spokes is: 
Ispokes = 8(1/12) m L^2 = (2/3) m L^2

The moment of inertia of the outer ring is given by: 
Iring = (1/2) mR^2

Since the ring is thin, its radius is equal to the radius of the wheel. The radius of the wheel is L/2, so 
Iring = (1/2) m(L/2)^2 = (1/8) m L^2

Therefore, the total moment of inertia of the wagon wheel is: 
I = Ispokes + Iring = (2/3) m L^2 + (1/8) m L^2 = (17/24) m L^2

The correct answer choice is B.

Quiz 9: Rotation

A homogeneous solid cylinder of mass m,length L,and radius R rotates about an axis through point P,which is parallel to the cylinder axis.If the moment of inertia about the cylinder axis is mR²,the moment of inertia about the axis through P is

In the figure,R₁ = R₂ and cm is the center of mass.The rotational inertia about an axis through point P₁ is I₁,the rotational inertia about an axis through point P₂ is I₂,and the rotational inertia about an axis through the cm is I_cm.The relationship among the moments is

A body of mass m is whirled at a constant angular velocity on the end of a string of length R.To double the kinetic energy of the body as it whirls while maintaining the angular velocity,the length of the string must be changed to

A ball of mass m₁,connected to another mass m₂ by a string,is whirled at a constant speed in a horizontal circle of radius R equal to 0.800m.If the mass m₂ = 5.00 kg,the kinetic energy of the ball is

To increase the moment of inertia of a body about an axis,you must

The amount of work done on a rotating body can be expressed in terms of the product of

The disc brake of a high performance car is often made of carbon fiber instead of iron,thereby reducing the mass.If the mass of a carbon disc is that of iron for the same shape and size,by what factor is the moment of inertia changed?

The rotational inertia of an object about an axis depends on the

The moment of inertia of a slim rod about a transverse axis through one end is mL²/3,where m is the mass of the rod and L is its length.The moment of inertia of a 0.24-kg meterstick about a transverse axis through its center is

A stone of mass 10 kg has a rotational inertia of 2.4 kg · m² about an axis A parallel to an axis through the center of mass.If axis A is 0.20m from the center of mass axis,the rotational inertia about the center of mass axis is

Four 50-g point masses are at the corners of a square with 20-cm sides.What is the moment of inertia of this system about an axis perpendicular to the plane of the square and passing through its center?

A hoop of mass 50 kg rolls without slipping.If the center-of-mass of the hoop has a translational speed of 4.0 m/s,the total kinetic energy of the hoop is

A 7.00-kg mass and a 4.00-kg mass are mounted on a spindle free to turn about the x axis as shown.Assume the mass of the arms and the spindle to be negligible.The rotational inertia of this system is approximately

The moment of inertia of a slim rod of mass m and length L about a transverse axis through one end is mL²/3.The moment of inertia of such a rod about a transverse axis through the rod at a distance L/3 from one end is

A spherical shell and a solid sphere (both of the same total mass) are rotating about a diameter at the same angular frequency and with the same rotational kinetic energy.What is the ratio of the spherical shell radius to the solid sphere (i.e.,R_{spherical shell}/R_{solid sphere}) ?

The moment of inertia of a set of dumbbells,considered as two mass points m separated by a distance 2L about the axis AA,is

A wagon wheel consists of 8 spokes of uniform diameter,each of mass m_s and length L cm.The outer ring has a mass m_ring.What is the moment of inertia of the wheel? Assume that each spoke extends from the center to the other ring and the ring is of negligible thickness.

Quiz 9: Rotation

A homogeneous solid cylinder of mass m,length L,and radius R rotates about an axis through point P,which is parallel to the cylinder axis.If the moment of inertia about the cylinder axis is mR2,the moment of inertia about the axis through P is

In the figure,R1 = R2 and cm is the center of mass.The rotational inertia about an axis through point P1 is I1,the rotational inertia about an axis through point P2 is I2,and the rotational inertia about an axis through the cm is Icm.The relationship among the moments is

A body of mass m is whirled at a constant angular velocity on the end of a string of length R.To double the kinetic energy of the body as it whirls while maintaining the angular velocity,the length of the string must be changed to

A ball of mass m1,connected to another mass m2 by a string,is whirled at a constant speed in a horizontal circle of radius R equal to 0.800m.If the mass m2 = 5.00 kg,the kinetic energy of the ball is

To increase the moment of inertia of a body about an axis,you must

The amount of work done on a rotating body can be expressed in terms of the product of

The disc brake of a high performance car is often made of carbon fiber instead of iron,thereby reducing the mass.If the mass of a carbon disc is that of iron for the same shape and size,by what factor is the moment of inertia changed?

The rotational inertia of an object about an axis depends on the

The moment of inertia of a slim rod about a transverse axis through one end is mL2/3,where m is the mass of the rod and L is its length.The moment of inertia of a 0.24-kg meterstick about a transverse axis through its center is

A stone of mass 10 kg has a rotational inertia of 2.4 kg · m2 about an axis A parallel to an axis through the center of mass.If axis A is 0.20m from the center of mass axis,the rotational inertia about the center of mass axis is

Four 50-g point masses are at the corners of a square with 20-cm sides.What is the moment of inertia of this system about an axis perpendicular to the plane of the square and passing through its center?

A hoop of mass 50 kg rolls without slipping.If the center-of-mass of the hoop has a translational speed of 4.0 m/s,the total kinetic energy of the hoop is

A 7.00-kg mass and a 4.00-kg mass are mounted on a spindle free to turn about the x axis as shown.Assume the mass of the arms and the spindle to be negligible.The rotational inertia of this system is approximately

The moment of inertia of a slim rod of mass m and length L about a transverse axis through one end is mL2/3.The moment of inertia of such a rod about a transverse axis through the rod at a distance L/3 from one end is

A spherical shell and a solid sphere (both of the same total mass) are rotating about a diameter at the same angular frequency and with the same rotational kinetic energy.What is the ratio of the spherical shell radius to the solid sphere (i.e.,Rspherical shell/Rsolid sphere) ?

A solid sphere of radius r = 5 cm and mass m = 0.2 kg rolls in a groove as shown.The angle, θ\thetaθ ,between the points of contact with the groove is 120 ∘\circ∘ .If the sphere has a linear velocity of 10 mm/s,what is the rotational kinetic energy?

The moment of inertia of a set of dumbbells,considered as two mass points m separated by a distance 2L about the axis AA,is

A wagon wheel consists of 8 spokes of uniform diameter,each of mass ms and length L cm.The outer ring has a mass mring.What is the moment of inertia of the wheel? Assume that each spoke extends from the center to the other ring and the ring is of negligible thickness.

A homogeneous solid cylinder of mass m,length L,and radius R rotates about an axis through point P,which is parallel to the cylinder axis.If the moment of inertia about the cylinder axis is mR²,the moment of inertia about the axis through P is

In the figure,R₁ = R₂ and cm is the center of mass.The rotational inertia about an axis through point P₁ is I₁,the rotational inertia about an axis through point P₂ is I₂,and the rotational inertia about an axis through the cm is I_cm.The relationship among the moments is

A ball of mass m₁,connected to another mass m₂ by a string,is whirled at a constant speed in a horizontal circle of radius R equal to 0.800m.If the mass m₂ = 5.00 kg,the kinetic energy of the ball is

The moment of inertia of a slim rod about a transverse axis through one end is mL²/3,where m is the mass of the rod and L is its length.The moment of inertia of a 0.24-kg meterstick about a transverse axis through its center is

A stone of mass 10 kg has a rotational inertia of 2.4 kg · m² about an axis A parallel to an axis through the center of mass.If axis A is 0.20m from the center of mass axis,the rotational inertia about the center of mass axis is

The moment of inertia of a slim rod of mass m and length L about a transverse axis through one end is mL²/3.The moment of inertia of such a rod about a transverse axis through the rod at a distance L/3 from one end is

A spherical shell and a solid sphere (both of the same total mass) are rotating about a diameter at the same angular frequency and with the same rotational kinetic energy.What is the ratio of the spherical shell radius to the solid sphere (i.e.,R_{spherical shell}/R_{solid sphere}) ?

A wagon wheel consists of 8 spokes of uniform diameter,each of mass m_s and length L cm.The outer ring has a mass m_ring.What is the moment of inertia of the wheel? Assume that each spoke extends from the center to the other ring and the ring is of negligible thickness.