Question 1

A wheel of diameter 3.0 cm has a 4.0 m cord wrapped around its periphery.Starting from rest, the wheel is given a constant angular acceleration of 2 rad/s².The cord will unwind in:

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Question 2

The rotational inertia of a thin cylindrical shell of mass M, radius R, and length L about its central axis (X - X')is:

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Question 3

Two wheels are identical but wheel B is spinning with twice the angular speed of wheel A.The ratio of the magnitude of the radial acceleration of a point on the rim of B to the magnitude of the radial acceleration of a point on the rim of A is:

Accepted Answer

The radial (centripetal) acceleration $a_r$ is given by $a_r = \omega^2 r$, where $\omega$ is the angular speed and $r$ is the radius of the wheel. Since both wheels are identical, their radii are the same. If wheel B spins with twice the angular speed of wheel A ($\omega_B = 2\omega_A$), then the radial acceleration of B is $a_{rB} = (2\omega_A)^2 r = 4\omega_A^2 r$, which is 4 times the radial acceleration of A ($a_{rA} = \omega_A^2 r$). Therefore, the ratio is 4:1.

Question 4

A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 *10^-3 kg.m² is suspended from the ceiling.A rope passes over it with a 2.0-kg block attached to one end and a 4.0-kg block attached to the other.The rope does not slip on the pulley.At any instant after the blocks start moving the object with the greatest kinetic energy is:

Accepted Answer

The pulley has the greatest rotational kinetic energy, which is given by 1/2 I w^2, where I is the rotational inertia and w is the angular velocity. The pulley has the greatest rotational inertia, so it will have the greatest rotational kinetic energy.

Question 5

Four identical particles, each with mass m, are arranged in the x, y plane as shown.They are connected by light sticks to form a rigid body.If m = 2.0 kg and a = 1.0 m, the rotational inertia of this array about the y-axis is:

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Question 6

Three identical balls are tied by light strings to the same rod and rotate around it, as shown below.Rank the balls according to their rotational inertia, least to greatest.

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Question 7

For a wheel spinning on an axis through its center, the ratio of the radial acceleration of a point on the rim to the radial acceleration of a point halfway between the center and the rim is:

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The radial acceleration of a point on the rim is a = v^2/r, where v is the tangential velocity and r is the radius. The tangential velocity is proportional to the distance from the axis, so v_rim = r*w, and v_center = (r/2)*w. Therefore, the ratio of the radial acceleration at the rim to the radial acceleration at the center is (v_rim)^2/(v_center)^2 = (r*w)^2/[(r/2)*w]^2 = 4, which means that the answer is choice B.

Question 8

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s².The tangential acceleration of a point on its rim is:

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Question 9

Wrapping paper is being unwrapped from a 5.0-cm radius tube, free to rotate on its axis.If it is pulled at the constant rate of 10 cm/s and does not slip on the tube, the angular velocity of the tube is:

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The angular velocity ($\omega$) can be calculated using the formula $\omega = v/r$, where $v$ is the linear velocity (10 cm/s) and $r$ is the radius (5.0 cm). Thus, $\omega = 10 \, \text{cm/s} / 5.0 \, \text{cm} = 2.0 \, \text{rad/s}$.

Question 10

Three balls, with masses of 3M, 2M, and M, are fastened to a massless rod of length L as shown.The rotational inertia about the left end of the rod is:

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Question 11

For a wheel spinning on an axis through its center, the ratio of the tangential acceleration of a point on the rim to the tangential acceleration of a point halfway between the center and the rim is:

Accepted Answer

The tangential acceleration of a point on the rim is given by $a_t=r\alpha$, where $r$ is the radius of the wheel and $\alpha$ is the angular acceleration. The point halfway between the center and the rim has a radius of $r/2$, so its tangential acceleration is $(r/2)\alpha$. Therefore, the ratio of the tangential acceleration of a point on the rim to the tangential acceleration of a point halfway between the center and the rim is:

$\frac{a_t}{a_{t/2}}=\frac{r\alpha}{(r/2)\alpha}=2$

So the correct answer is B.

Question 12

A wheel starts from rest and spins with a constant angular acceleration.As time goes on the acceleration vector for a point on the rim:

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The acceleration vector for a point on the rim of a wheel that starts from rest and spins with a constant angular acceleration includes both the tangential acceleration (due to the change in speed) and the radial (centripetal) acceleration (due to the change in direction). While the tangential acceleration remains constant due to the constant angular acceleration, the radial acceleration increases in magnitude because it is proportional to the square of the tangential speed, which increases as the wheel spins faster. Thus, the overall acceleration vector increases in magnitude and becomes more nearly radial as the tangential speed increases.

Question 13

A particle moves in a circular path of radius 0.10 m with a constant angular speed of 5 rev/s.The acceleration of the particle is:

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Question 14

The figure shows a cylinder of radius 0.7 m rotating about its axis at 10 rad/s.The speed of the point P is:

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The speed of a point on a rotating cylinder is equal to the product of the radius and the angular velocity. Therefore, the speed of point P is 0.7m x 10 rad/s = 7.0m/s. Option A is the correct answer.

Question 15

For a wheel spinning with constant angular acceleration on an axis through its center, the ratio of the speed of a point on the rim to the speed of a point halfway between the center and the rim is:

Accepted Answer

Let ω be the angular velocity, α be the angular acceleration, v1 be the velocity of a point on the rim, and v2 be the velocity of a point halfway between the center and the rim. Then we have:

v1 = ωR
v2 = ω(R/2)

The tangential acceleration of a point on the rim is a1 = Rα, and the tangential acceleration of a point halfway between the center and the rim is a2 = (R/2)α.

By the formula for tangential acceleration, we have

a1 = Rα = dv1/dt = R(dω/dt)
a2 = (R/2)α = dv2/dt = (R/2)(dω/dt)

Dividing these two equations by each other,

a1/a2 = (R/R/2) = 2

But the tangential acceleration is also equal to the ratio of the speeds squared:

a1/a2 = v1^2/v2^2

So we have:

v1^2/v2^2 = 2

Taking the square root of both sides,

v1/v2 = sqrt(2)

Therefore, the ratio of the speed of a point on the rim to the speed of a point halfway between the center and the rim is 2.

Question 16

A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 * 10^-3 kg∙m² is suspended from the ceiling.A rope passes over it with a 2.0-kg block attached to one end and a 4.0-kg block attached to the other.The rope does not slip on the pulley.When the velocity of the heavier block is 2.0 m/s the total kinetic energy of the pulley and blocks is:

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The total kinetic energy (KE) of the system includes the translational kinetic energy of the two blocks and the rotational kinetic energy of the pulley. The translational kinetic energy (KE_trans) for each block is given by $KE_{trans} = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. The rotational kinetic energy (KE_rot) of the pulley is given by $KE_{rot} = \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia (rotational inertia) and $\omega$ is the angular velocity.Given:- Mass of block 1 (m1) = 2.0 kg- Mass of block 2 (m2) = 4.0 kg- Velocity of blocks (v) = 2.0 m/s- Radius of pulley (r) = 0.03 m (3.0 cm)- Rotational inertia of pulley (I) = $4.5 \times 10^{-3}$ kg·m²First, calculate the translational kinetic energy for both blocks:$$KE_{trans1} = \frac{1}{2}m_1v^2 = \frac{1}{2} \times 2.0 \times (2.0)^2 = 4.0 \, \text{J}$$$$KE_{trans2} = \frac{1}{2}m_2v^2 = \frac{1}{2} \times 4.0 \times (2.0)^2 = 8.0 \, \text{J}$$To find the rotational kinetic energy of the pulley, we first need to find the angular velocity ($\omega$). The linear velocity (v) is related to angular velocity ($\omega$) by $v = r\omega$, so:$$\omega = \frac{v}{r} = \frac{2.0}{0.03} = \frac{200}{3} \, \text{rad/s}$$Now, calculate the rotational kinetic energy of the pulley:$$KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 4.5 \times 10^{-3} \times \left(\frac{200}{3}\right)^2$$$$KE_{rot} = \frac{1}{2} \times 4.5 \times 10^{-3} \times \frac{40000}{9} = 4.5 \times 10^{-3} \times \frac{40000}{18}$$$$KE_{rot} = 10 \, \text{J}$$Finally, the total kinetic energy of the system is the sum of the translational kinetic energies of both blocks and the rotational kinetic energy of the pulley:$$KE_{total} = KE_{trans1} + KE_{trans2} + KE_{rot} = 4.0 + 8.0 + 10.0 = 22.0 \, \text{J}$$

Question 17

The rotational inertia of a wheel about its axle does not depend upon its:

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The rotational inertia of a wheel about its axle depends only on its mass and distribution of mass, as well as the distance between the axis of rotation and the mass distribution (i.e. the radius). The speed of rotation is not a factor in determining the rotational inertia.

Question 18

The magnitude of the acceleration of a point on a spinning wheel is increased by a factor of 4 if:

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Question 19

String is wrapped around the periphery of a 5.0-cm radius cylinder, free to rotate on its axis.The string is pulled straight out at a constant rate of 10 cm/s and does not slip on the cylinder.As each small segment of string leaves the cylinder, the segment's acceleration changes by:

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Question 20

A car travels north at constant velocity.It goes over a piece of mud which sticks to the tire.The initial acceleration of the mud, as it leaves the ground, is:

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The initial acceleration of the mud as it leaves the ground is vertically upward due to the normal force exerted by the ground on the mud, overcoming gravity and causing it to stick to the tire and move upwards.

Quiz 10: Rotation

A wheel of diameter 3.0 cm has a 4.0 m cord wrapped around its periphery.Starting from rest, the wheel is given a constant angular acceleration of 2 rad/s².The cord will unwind in:

The rotational inertia of a thin cylindrical shell of mass M, radius R, and length L about its central axis (X - X') is:

Two wheels are identical but wheel B is spinning with twice the angular speed of wheel A.The ratio of the magnitude of the radial acceleration of a point on the rim of B to the magnitude of the radial acceleration of a point on the rim of A is:

Four identical particles, each with mass m, are arranged in the x, y plane as shown.They are connected by light sticks to form a rigid body.If m = 2.0 kg and a = 1.0 m, the rotational inertia of this array about the y-axis is:

Three identical balls are tied by light strings to the same rod and rotate around it, as shown below.Rank the balls according to their rotational inertia, least to greatest.

For a wheel spinning on an axis through its center, the ratio of the radial acceleration of a point on the rim to the radial acceleration of a point halfway between the center and the rim is:

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s².The tangential acceleration of a point on its rim is:

Wrapping paper is being unwrapped from a 5.0-cm radius tube, free to rotate on its axis.If it is pulled at the constant rate of 10 cm/s and does not slip on the tube, the angular velocity of the tube is:

Three balls, with masses of 3M, 2M, and M, are fastened to a massless rod of length L as shown.The rotational inertia about the left end of the rod is:

For a wheel spinning on an axis through its center, the ratio of the tangential acceleration of a point on the rim to the tangential acceleration of a point halfway between the center and the rim is:

A wheel starts from rest and spins with a constant angular acceleration.As time goes on the acceleration vector for a point on the rim:

A particle moves in a circular path of radius 0.10 m with a constant angular speed of 5 rev/s.The acceleration of the particle is:

The figure shows a cylinder of radius 0.7 m rotating about its axis at 10 rad/s.The speed of the point P is: $The figure shows a cylinder of radius 0.7 m rotating about its axis at 10 rad/s.The speed of the point P is: A) 7.0 m/s B) 14 \pi rad/s C) 7 \pi rad/s D) 0.70 m/s E) none of these$

For a wheel spinning with constant angular acceleration on an axis through its center, the ratio of the speed of a point on the rim to the speed of a point halfway between the center and the rim is:

The rotational inertia of a wheel about its axle does not depend upon its:

The magnitude of the acceleration of a point on a spinning wheel is increased by a factor of 4 if:

String is wrapped around the periphery of a 5.0-cm radius cylinder, free to rotate on its axis.The string is pulled straight out at a constant rate of 10 cm/s and does not slip on the cylinder.As each small segment of string leaves the cylinder, the segment's acceleration changes by:

A car travels north at constant velocity.It goes over a piece of mud which sticks to the tire.The initial acceleration of the mud, as it leaves the ground, is:

Quiz 10: Rotation

A wheel of diameter 3.0 cm has a 4.0 m cord wrapped around its periphery.Starting from rest, the wheel is given a constant angular acceleration of 2 rad/s2.The cord will unwind in:

The rotational inertia of a thin cylindrical shell of mass M, radius R, and length L about its central axis (X - X') is:

Two wheels are identical but wheel B is spinning with twice the angular speed of wheel A.The ratio of the magnitude of the radial acceleration of a point on the rim of B to the magnitude of the radial acceleration of a point on the rim of A is:

Four identical particles, each with mass m, are arranged in the x, y plane as shown.They are connected by light sticks to form a rigid body.If m = 2.0 kg and a = 1.0 m, the rotational inertia of this array about the y-axis is:

Three identical balls are tied by light strings to the same rod and rotate around it, as shown below.Rank the balls according to their rotational inertia, least to greatest.

For a wheel spinning on an axis through its center, the ratio of the radial acceleration of a point on the rim to the radial acceleration of a point halfway between the center and the rim is:

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2.The tangential acceleration of a point on its rim is:

Wrapping paper is being unwrapped from a 5.0-cm radius tube, free to rotate on its axis.If it is pulled at the constant rate of 10 cm/s and does not slip on the tube, the angular velocity of the tube is:

Three balls, with masses of 3M, 2M, and M, are fastened to a massless rod of length L as shown.The rotational inertia about the left end of the rod is:

For a wheel spinning on an axis through its center, the ratio of the tangential acceleration of a point on the rim to the tangential acceleration of a point halfway between the center and the rim is:

A wheel starts from rest and spins with a constant angular acceleration.As time goes on the acceleration vector for a point on the rim:

A particle moves in a circular path of radius 0.10 m with a constant angular speed of 5 rev/s.The acceleration of the particle is:

The figure shows a cylinder of radius 0.7 m rotating about its axis at 10 rad/s.The speed of the point P is:

For a wheel spinning with constant angular acceleration on an axis through its center, the ratio of the speed of a point on the rim to the speed of a point halfway between the center and the rim is:

The rotational inertia of a wheel about its axle does not depend upon its:

The magnitude of the acceleration of a point on a spinning wheel is increased by a factor of 4 if:

String is wrapped around the periphery of a 5.0-cm radius cylinder, free to rotate on its axis.The string is pulled straight out at a constant rate of 10 cm/s and does not slip on the cylinder.As each small segment of string leaves the cylinder, the segment's acceleration changes by:

A car travels north at constant velocity.It goes over a piece of mud which sticks to the tire.The initial acceleration of the mud, as it leaves the ground, is:

A wheel of diameter 3.0 cm has a 4.0 m cord wrapped around its periphery.Starting from rest, the wheel is given a constant angular acceleration of 2 rad/s².The cord will unwind in:

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s².The tangential acceleration of a point on its rim is:

The figure shows a cylinder of radius 0.7 m rotating about its axis at 10 rad/s.The speed of the point P is: $The figure shows a cylinder of radius 0.7 m rotating about its axis at 10 rad/s.The speed of the point P is: A) 7.0 m/s B) 14 \pi rad/s C) 7 \pi rad/s D) 0.70 m/s E) none of these$