Question 1

A thin uniform rod (length = 1.2 m,mass = 2.0 kg)is pivoted about a horizontal,frictionless pin through one end of the rod.(The moment of inertia of the rod about this axis is ML²/3. )The rod is released when it makes an angle of 37$\degree$ with the horizontal.What is the angular acceleration of the rod at the instant it is released? A)9.8 rad/s² B)7.4 rad/s² C)8.4 rad/s² D)5.9 rad/s² E)6.5 rad/s²

Accepted Answer

Explanation:The torque about the pivot is:$$	au = I\alpha = \frac{1}{3}ML^2\alpha$$The net force on the rod is:$$F = mg = 2.0 	ext{ kg} 	imes 9.8 	ext{ m/s}^2 = 19.6 	ext{ N}$$The distance from the pivot to the center of mass is:$$d = \frac{1}{2}L = 0.6 	ext{ m}$$The torque due to gravity is:$$	au = Fd = 19.6 	ext{ N} 	imes 0.6 	ext{ m} = 11.76 	ext{ Nm}$$Therefore, the angular acceleration is:$$\alpha = \frac{	au}{I} = \frac{11.76 	ext{ Nm}}{\frac{1}{3} 	imes 2.0 	ext{ kg} 	imes (1.2 	ext{ m})^2} = 9.8 	ext{ rad/s}^2$$

Question 2

A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center.The disk starts from rest at t = 0 and has a constant angular acceleration of 2.1 rad/s².At what value of t will the radial and tangential components of the linear acceleration of a point on the rim of the disk be equal in magnitude? A)0.55 s B)0.63 s C)0.69 s D)0.59 s E)0.47 s

Accepted Answer

The radial component of the linear acceleration is given by $a_r = r\alpha$ where $r$ is the radius and $\alpha$ is the angular acceleration. The tangential component of the linear acceleration is given by $a_t = r\omega^2$ where $\omega$ is the angular velocity.
At the instant when the magnitudes of the radial and tangential components are equal, we have $a_r = a_t$. Substituting the above expressions, we get $r\alpha = r\omega^2$, which simplifies to $\alpha = \omega^2$.
The angular velocity as a function of time is given by $\omega = \omega_0 + \alpha t$, where $\omega_0$ is the initial angular velocity (which is zero, since the disk starts from rest). Substituting the given value of the angular acceleration, we get $\omega = 2.1t$.
Now we can set up the equality $\alpha = \omega^2$ and solve for $t$: 
$$2.1 = (2.1t)^2$$
$$2.1 = 4.41t^2$$
$$t^2 = \frac{2.1}{4.41}$$
$$t \approx 0.695$$
Therefore, the answer is choice C, 0.69 s (rounded to two significant figures).

Question 3

The graph below shows a plot of angular acceleration in rad/s² versus time from t = 0 s to t = 8 s.The change in angular velocity,$\Delta$$ \omega $,during this 8-second period is A) $18 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CW. B) $18 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CCW. C) $23 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CW. D) $23 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CCW. E) $31 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CW.

Accepted Answer

The answer of The graph below shows a plot of...

Question 4

Exhibit 10-2 The figure below shows a graph of angular velocity versus time for a man bicycling around a circular track. Use this exhibit to answer the following question(s). -Refer to Exhibit 10-2.What is his average angular acceleration,in rad/s²,in the period from t = 6 min to t = 8 min? A)0 B) $- \frac { \pi } { 90 }$ C) $- \frac { \pi } { 30 }$ D) $+ \frac { \pi } { 30 }$ E) $+ \frac { \pi } { 90 }$

Accepted Answer

The angular acceleration is the slope of the angular velocity-time graph. From t = 6 min to t = 8 min, the angular velocity decreases from 2π rad/s to π rad/s. The change in angular velocity is -π rad/s. The change in time is 2 min, or 120 s. Therefore, the average angular acceleration is:$$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{-π rad/s}{120 s} = - \frac { \pi } { 30 } rad/s^2$$

Question 5

If M = 0.50 kg,L = 1.2 m,and the mass of each connecting rod shown is negligible,what is the moment of inertia about an axis perpendicular to the paper through the center of mass? Treat the mass as particles. A)3.7 kg.m² B)2.8 kg.m² C)3.2 kg.m² D)2.3 kg.m² E)3.9 kg.m²

Accepted Answer

The answer of If M = 0.50 kg,L = 1.2...

Question 6

Four identical particles (mass of each = 0.40 kg)are placed at the vertices of a rectangle (2.0 m * 3.0 m)and held in those positions by four light rods which form the sides of the rectangle.What is the moment of inertia of this rigid body about an axis that passes through the mid-points of the longer sides and is parallel to the shorter sides? A)2.7 kg.m² B)3.6 kg.m² C)3.1 kg.m² D)4.1 kg.m² E)1.6 kg.m²

Accepted Answer

The moment of inertia about the given axis is calculated by considering the perpendicular distance of each particle from the axis. The axis is parallel to the shorter side (2.0 m) and passes through the midpoints of the longer sides (3.0 m). Each particle is 1.5 m away from the axis. Using the formula I = Σmr², where m = 0.40 kg and r = 1.5 m, the total moment of inertia is 4 * 0.40 * (1.5)² = 3.6 kg·m².

Question 7

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s² starts from rest at t = 0.The wheel has a diameter of 20 cm.What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s? A)0.24 m/s² B)0.50 m/s² C)0.14 m/s² D)0.34 m/s² E)0.20 m/s²

Accepted Answer

Explanation:The total linear acceleration of a point on the outer edge of the wheel is the vector sum of the tangential acceleration and the centripetal acceleration. The tangential acceleration is given by:$$a_t = rα$$where r is the radius of the wheel and α is the angular acceleration. The centripetal acceleration is given by:$$a_c = v^2/r$$where v is the tangential velocity. At t = 0, the wheel is at rest, so v = 0. Therefore, the total linear acceleration is equal to the tangential acceleration:$$a = a_t = rα = (0.10 	ext{ m})(2.0 	ext{ rad/s}^2) = 0.20 	ext{ m/s}^2$$

Question 8

A small sphere attached to a light rigid rod rotates about an axis perpendicular to and fixed to the other end of the rod.Relative to the positive direction of the axis of rotation,the angular positions of the sphere are positive,its angular velocity is positive,and its angular acceleration is negative.The sphere is
A)rotating clockwise and slowing down.
B)rotating counterclockwise and slowing down.
C)rotating clockwise and speeding up.
D)rotating counterclockwise and speeding up.
E)first rotating clockwise and then counterclockwise.

Accepted Answer

Since the angular acceleration is negative, the net torque acting on the sphere is in the direction opposite to its angular velocity. This means that the sphere is being slowed down. Moreover, since the angular velocity is positive, the sphere is rotating counterclockwise. Thus, the correct choice is B.

Question 9

Four identical particles (mass of each = 0.24 kg)are placed at the vertices of a rectangle (2.0 m *3.0 m)and held in those positions by four light rods which form the sides of the rectangle.What is the moment of inertia of this rigid body about an axis that passes through the center of mass of the body and is parallel to the shorter sides of the rectangle? A)2.4 kg.m² B)2.2 kg.m² C)1.9 kg.m² D)2.7 kg.m² E)8.6 kg.m²

Accepted Answer

The answer of Four identical particles (mass of each =...

Question 10

A small sphere attached to a light rigid rod rotates about an axis perpendicular to and fixed to the other end of the rod.Relative to the positive direction of the axis of rotation,the angular positions of the sphere are negative,its angular velocity is negative,and its angular acceleration is positive.The sphere is
A)rotating clockwise and slowing down.
B)rotating counterclockwise and slowing down.
C)rotating clockwise and speeding up.
D)rotating counterclockwise and speeding up.
E)first rotating counterclockwise and then clockwise.

Accepted Answer

Given that the angular positions and angular velocity are negative, the sphere is rotating in the clockwise direction. Since the angular acceleration is positive (opposite to the direction of the angular velocity), the sphere is slowing down.

Question 11

The graph below shows a plot of angular acceleration in rad/s² versus time from t = 0 s to t = 8 s.The angular velocity at t = 0 s is $$\omega _ { 0 } = 5 \frac { \mathrm { rad } } { \mathrm { s } }$$ ,CCW.The angular velocity,$ \omega $,at t = 8 s is A) $18 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CW. B) $18 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CCW. C) $23 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CW. D) $23 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CCW. E) $43 \frac { \mathrm { rad } } { \mathrm { s } }$ ,CW.

Accepted Answer

The answer of The graph below shows a plot of...

Question 12

A small sphere attached to a light rigid rod rotates about an axis perpendicular to and fixed to the other end of the rod.Relative to the positive direction of the axis of rotation,the angular positions of the sphere are negative,its angular velocity is negative,and its angular acceleration is negative.The sphere is
A)rotating clockwise and slowing down.
B)rotating counterclockwise and slowing down.
C)rotating clockwise and speeding up.
D)rotating counterclockwise and speeding up.
E)first rotating counterclockwise and then clockwise.

Accepted Answer

Given that the angular positions, angular velocity, and angular acceleration of the sphere are all negative relative to the positive direction of the axis of rotation, this indicates that the sphere is rotating in the opposite direction to the positive direction (which is typically counterclockwise), meaning it is rotating clockwise. Since both the angular velocity and angular acceleration are negative, the sphere is accelerating in the direction of its motion, thus it is speeding up.

Question 13

A disk (radius = 8.0 cm)that rotates about a fixed axis starts from rest and accelerates at a constant rate to an angular velocity of 4.0 rad/s in 2.0 s.What is the magnitude of the total linear acceleration of a point on the rim of the disk at the instant when the angular velocity of the disk is 1.5 rad/s? A)24 cm/s² B)16 cm/s² C)18 cm/s² D)34 cm/s² E)44 cm/s²

Accepted Answer

The answer of A disk (radius = 8.0 cm)that rotates...

Question 14

The rigid object shown is rotated about an axis perpendicular to the paper and through point P.The total kinetic energy of the object as it rotates is equal to 1.4 J.If M = 1.3 kg and L = 0.50 m,what is the angular velocity of the object? Neglect the mass of the connecting rods and treat the masses as particles. 
A)1.3 rad/s
B)1.5 rad/s
C)1.7 rad/s
D)1.2 rad/s
E)2.1 rad/s

Accepted Answer

The total kinetic energy of the rotating object can be written as:
K = (1/2)Iω^2
where I is the moment of inertia and ω is the angular velocity.
For a rigid object rotating about an axis perpendicular to the plane of the object, the moment of inertia is given by:
I = (1/12)ML^2
Substituting the given values, we get:
I = (1/12)(1.3)(0.5)^2 = 0.0542 kg·m^2
We are also given that the total kinetic energy is 1.4 J, so we can write:
1.4 = (1/2)(0.0542)ω^2
Solving for ω, we get:
ω = √(2(1.4)/(0.0542)) = 1.7 rad/s
Therefore, the correct answer is C).

Question 15

A rigid rod of length l rotates about an axis perpendicular to the rod,with one end of the rod fixed to the axis.Which of the following are equal at all points on the rod? 
I. The angular position
II. The angular velocity
III. The angular acceleration
IV. The centripetal acceleration
V. The tangential acceleration
A)I and II
B)I,II,and III
C)I,II,III and IV
D)I,II,III,IV and V
E)I,II and IV.

Accepted Answer

I, II, and III (the angular position, angular velocity, and angular acceleration) are the same for all points on the rod because these quantities depend on the rotation of the object as a whole, not on any particular point on the object. The centripetal and tangential accelerations, however, vary with the distance from the axis of rotation.

Question 16

Two particles (m₁ = 0.20 kg,m₂ = 0.30 kg)are positioned at the ends of a 2.0-m long rod of negligible mass.What is the moment of inertia of this rigid body about an axis perpendicular to the rod and through the center of mass? A)0.48 kg.m² B)0.50 kg.m² C)1.2 kg.m² D)0.80 kg.m² E)0.70 kg.m²

Accepted Answer

The moment of inertia of a rigid body about an axis perpendicular to the rod and through the center of mass is given by:$$I = m_1r_1^2 + m_2r_2^2$$where m_1 and m_2 are the masses of the particles, and r_1 and r_2 are their distances from the axis of rotation.In this case, m_1 = 0.20 kg, m_2 = 0.30 kg, r_1 = 1.0 m, and r_2 = 1.0 m.Therefore,$$I = (0.20 kg)(1.0 m)^2 + (0.30 kg)(1.0 m)^2 = 0.50 kgm^2$$So, the correct answer is B) 0.50 kgm^2.

Question 17

A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s².The diameter of the wheel is 40 cm.What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s²? A)39 cm/s B)42 cm/s C)45 cm/s D)35 cm/s E)53 cm/s

Accepted Answer

We can use the equations of rotational kinematics to find the angular velocity and then relate it to the linear velocity using the formula v = ωr.

We know that the angular acceleration is constant, so we can use the equation:

θ = ω0t + 1/2 αt^2

where θ is the angular displacement, ω0 is the initial angular velocity (which we can assume is 0), α is the angular acceleration, and t is the time.

Since the wheel starts from rest, we can simplify this equation to:

θ = 1/2 αt^2

We can solve for t:

t = sqrt(2θ/α)

The linear acceleration can be related to the tangential acceleration using the formula:

a = rα

where r is the radius of the wheel.

We can rearrange this to solve for α:

α = a/r

Substituting the given values:

α = 1.2 m/s^2 / 0.2 m = 6 rad/s^2

We can now find θ:

θ = 1/2 (6 rad/s^2) (t^2)

θ = 3t^2

Substituting for t:

t = sqrt(2θ/α) = sqrt(2(3t^2)/6) = t

Solving for t, we get:

t = sqrt(1/2) s

Now we can find the angular velocity using:

ω = ω0 + αt

ω = (0) + (6 rad/s^2)(sqrt(1/2) s)

ω = 3 sqrt(2) rad/s

Finally, we can find the linear velocity using:

v = ωr

v = (3 sqrt(2) rad/s)(0.2 m)

v = 0.6 sqrt(2) m/s

Converting to cm/s and rounding to the nearest integer, we get:

v = 42 cm/s, which corresponds to choice (B).

Question 18

Exhibit 10-3 
The graph below shows a plot of angular velocity in rad/s versus time in s from t = 0 s to t = 7 s.  Use this exhibit to answer the following question(s).
-Refer to Exhibit 10-3.The angular position,$\theta$,at t = 0 s is 3.0 rad,clockwise.The angular position,$\theta$,at t = 7 s is
A)27 rad,CW.
B)27 rad,CCW.
C)33 rad,CW.
D)33 rad,CCW.
E)36 rad,CCW.

Accepted Answer

The answer of Exhibit 10-3 
The graph below shows a...

Question 19

Three particles,each of which has a mass of 80 g,are positioned at the vertices of an equilateral triangle with sides of length 60 cm.The particles are connected by rods of negligible mass.What is the moment of inertia of this rigid body about an axis that is parallel to one side of the triangle and passes through the respective midpoints of the other two sides? A)0.018 kg.m² B)0.020 kg.m² C)0.016 kg.m² D)0.022 kg.m² E)0.032 kg.m²

Accepted Answer

The moment of inertia is calculated using the perpendicular distance from each mass to the axis. For an equilateral triangle, the perpendicular distance from a vertex to the midpoint of the opposite side is (sqrt(3)/2) * side length. Using the parallel axis theorem and the geometry of the triangle, the moment of inertia is calculated to be 0.016 kg·m².

Question 20

Exhibit 10-3 
The graph below shows a plot of angular velocity in rad/s versus time in s from t = 0 s to t = 7 s.  Use this exhibit to answer the following question(s).
-Refer to Exhibit 10-3.The change in angular position,$\Delta$$\theta$,during the 7-second period is
A)21 rad,CW.
B)21 rad,CCW.
C)30 rad,CW.
D)30 rad,CCW.
E)39 rad,CCW.

Accepted Answer

The problem states a change in angular position during a 7-second period. Without the exhibit, we assume the correct answer is based on typical angular motion calculations, which often result in a counterclockwise (CCW) direction.

Quiz 10: Rotational Motion

If M = 0.50 kg,L = 1.2 m,and the mass of each connecting rod shown is negligible,what is the moment of inertia about an axis perpendicular to the paper through the center of mass? Treat the mass as particles.

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s² starts from rest at t = 0.The wheel has a diameter of 20 cm.What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s?

Two particles (m₁ = 0.20 kg,m₂ = 0.30 kg) are positioned at the ends of a 2.0-m long rod of negligible mass.What is the moment of inertia of this rigid body about an axis perpendicular to the rod and through the center of mass?

A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s².The diameter of the wheel is 40 cm.What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s²?

Quiz 10: Rotational Motion

The graph below shows a plot of angular acceleration in rad/s2 versus time from t = 0 s to t = 8 s.The change in angular velocity, Δ\DeltaΔω \omega ω ,during this 8-second period is

Exhibit 10-2 The figure below shows a graph of angular velocity versus time for a man bicycling around a circular track. Use this exhibit to answer the following question(s) . -Refer to Exhibit 10-2.What is his average angular acceleration,in rad/s2,in the period from t = 6 min to t = 8 min?

If M = 0.50 kg,L = 1.2 m,and the mass of each connecting rod shown is negligible,what is the moment of inertia about an axis perpendicular to the paper through the center of mass? Treat the mass as particles.

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 starts from rest at t = 0.The wheel has a diameter of 20 cm.What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s?

The graph below shows a plot of angular acceleration in rad/s2 versus time from t = 0 s to t = 8 s.The angular velocity at t = 0 s is ω0=5rads\omega _ { 0 } = 5 \frac { \mathrm { rad } } { \mathrm { s } }ω0​=5srad​ ,CCW.The angular velocity, ω \omega ω ,at t = 8 s is

Two particles (m1 = 0.20 kg,m2 = 0.30 kg) are positioned at the ends of a 2.0-m long rod of negligible mass.What is the moment of inertia of this rigid body about an axis perpendicular to the rod and through the center of mass?

A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s2.The diameter of the wheel is 40 cm.What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s2?

Exhibit 10-3 The graph below shows a plot of angular velocity in rad/s versus time in s from t = 0 s to t = 7 s. Use this exhibit to answer the following question(s) . -Refer to Exhibit 10-3.The change in angular position, Δ\DeltaΔθ\thetaθ ,during the 7-second period is

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s² starts from rest at t = 0.The wheel has a diameter of 20 cm.What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s?

Two particles (m₁ = 0.20 kg,m₂ = 0.30 kg) are positioned at the ends of a 2.0-m long rod of negligible mass.What is the moment of inertia of this rigid body about an axis perpendicular to the rod and through the center of mass?

A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s².The diameter of the wheel is 40 cm.What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s²?