Question 1

The disk in a DVD player has an angular velocity of 8.0 rad/s when it is turned off. The disk comes to rest 2.5 s after being turned off. Through how many radians does the disk rotate after being turned off? Assume constant angular acceleration.
A)12 rad
B)8.0 rad
C)10 rad
D)16 rad
E)6.8 rad

Accepted Answer

We can use the formula $	heta=\omega_it+\frac{1}{2}\alpha t^2$, where $	heta$ is the rotation in radians, $\omega_i$ is the initial angular velocity, $\alpha$ is the angular acceleration (assumed to be constant), and $t$ is the time. Plugging in the given values, we get $	heta=(8.0	ext{ rad/s})(2.5	ext{ s})+\frac{1}{2}(0)(2.5	ext{ s})^2=20	ext{ rad}$. However, this is the total rotation from when the DVD player is turned off until it comes to rest. To find the rotation after being turned off, we need to subtract the rotation that occurs during the 2.5 s it takes to come to rest. The rotation during that time is $	heta=(8.0	ext{ rad/s})(2.5	ext{ s})=20	ext{ rad}$. Therefore, the rotation after being turned off is $20	ext{ rad}-20	ext{ rad}=0	ext{ rad}$. Thus, the correct answer is $\boxed{	ext{(C) }10	ext{ rad}}$.

Question 2

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

C

Question 3

A wheel starts from rest and rotates with a constant angular acceleration about a fixed axis. It completes the first revolution 6.0 s after it started. How long after it started will the wheel complete the second revolution?
A)9.9 s
B)7.8 s
C)8.5 s
D)9.2 s
E)6.4 s

Accepted Answer

From the problem:
time taken to complete first revolution = 6s
We need to find time taken to complete second revolution.
For the second revolution, the wheel will have to cover 2π radians (as it completes one revolution in 2π radians), since the angular acceleration is constant, the angular displacement will be proportional to square of time, therefore,
θ = 1/2 * α * t^2
where θ is angular displacement, α is angular acceleration and t is time taken.
For first revolution, θ = π radians, t = 6s. Therefore we can calculate the angular acceleration as:
π = 1/2 * α * (6s)^2, which gives α = π/18 rad/s^2.
Now, for second revolution, θ = 2π radians, α = π/18 rad/s^2. Therefore we can calculate the time taken as:
2π = 1/2 * π/18 * t^2, which gives t = 8.5s.
Therefore, the wheel will complete the second revolution 8.5 seconds after it started.

Question 4

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:$$F = mg\sin	heta$$The acceleration of the rod is:$$a = \frac{F}{m} = g\sin	heta$$The angular acceleration of the rod is:$$\alpha = \frac{a}{L} = \frac{g\sin	heta}{L}$$$$\alpha = \frac{(9.8 m/s^2)\sin37\degree}{1.2 m} = 7.4 rad/s^2$$

Question 5

A wheel (radius = 0.20 m) starts from rest and rotates with a constant angular acceleration of 2.0 rad/s2. At the instant when the angular velocity is equal to 1.2 rad/s, what is the magnitude of the total linear acceleration of a point on the rim of the wheel?
A)0.40 m/s2
B)0.29 m/s2
C)0.69 m/s2
D)0.49 m/s2
E)0.35 m/s2

Accepted Answer

The total linear acceleration of a point on the rim 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. The total linear acceleration is given by:$$a = √(a_t^2 + a_c^2)$$Substituting the given values, we get:$$a = √((0.20 m)(2.0 rad/s^2))^2 + ((1.2 rad/s)^2/(0.20 m))^2)$$$$a = 0.49 m/s^2$$

Question 6

Two vectors lying in the xz plane are given by the equations  and  . The value of  is:
A) 
B) 
C)7 
D)(-7 )
E)

Accepted Answer

The answer of Two vectors lying in the xz plane...

Question 7

A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval the angular velocity increases to 40 rad/s. Assume that the angular acceleration was constant during the 5.0-s interval. How many revolutions does the wheel turn through during the 5.0-s interval?
A)20 rev
B)24 rev
C)32 rev
D)28 rev
E)39 rev

Accepted Answer

The total number of revolutions can be found using the formula for angular displacement $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$, where $\omega_0$ is the initial angular velocity, $\alpha$ is the angular acceleration, and $t$ is the time. First, find the angular acceleration $\alpha = \frac{\Delta \omega}{\Delta t} = \frac{40 \text{ rad/s} - 20 \text{ rad/s}}{5.0 \text{ s}} = 4 \text{ rad/s}^2$. Then, calculate the angular displacement in radians: $\theta = 20 \text{ rad/s} \times 5.0 \text{ s} + \frac{1}{2} \times 4 \text{ rad/s}^2 \times (5.0 \text{ s})^2 = 100 + 50 = 150 \text{ rad}$. Convert this to revolutions: $150 \text{ rad} \times \frac{1 \text{ rev}}{2\pi \text{ rad}} \approx 24 \text{ rev}$.

Question 8

Two vectors lying in the xy plane are given by the equations  and  . The value of  is:
A)19 
B)(-11  )
C)(-19  )
D)11 
E)10

Accepted Answer

We can find the dot product of the two given vectors:
(11, 8) · (10, -5) = 11(10) + 8(-5) = 90
The magnitude of the first vector is:
√((11)² + (8)²) = √(185)
Therefore, the value of cosθ is:
cosθ = 90/(√(185)*√(125)) ≈ 0.79819
θ ≈ 37.326°
Finally, the scalar projection of the second vector onto the first vector is:
||v2||cosθ ≈ 19.

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 centre 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 wheel (radius = 12 cm) is mounted on a frictionless, horizontal axle that is perpendicular to the wheel and passes through the centre of mass of the wheel. A light cord wrapped around the wheel supports a 0.40-kg object. If released from rest with the string taut, the object is observed to fall with a downward acceleration of 3.0 m/s2. What is the moment of inertia (of the wheel) about the given axle? A)0.023 kg.m² B)0.013 kg.m² C)0.020 kg.m² D)0.016 kg.m² E)0.035 kg.m²

Accepted Answer

The net torque on the wheel is equal to the tension in the cord times the radius of the wheel. Using Newton's second law for rotation and the linear acceleration of the object, we can solve for the moment of inertia.

Question 11

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 turns through 2.4 revolutions during a 2.0-s time interval. What is the angular velocity at the end of this time interval?
A)9.5 rad/s
B)9.7 rad/s
C)9.3 rad/s
D)9.1 rad/s
E)8.8 rad/s

Accepted Answer

The angular velocity at the end of the time interval is given by:ωf = ωi + αtwhere ωi is the initial angular velocity, α is the angular acceleration, and t is the time interval.In this case, ωi = 0 rad/s, α = 2.0 rad/s2, and t = 2.0 s. Therefore,ωf = 0 rad/s + 2.0 rad/s2 * 2.0 s = 4.0 rad/sThe wheel then turns through 2.4 revolutions during the 2.0-s time interval, which is equivalent to 4.8π radians. Therefore, the angular velocity at the end of the time interval is:ωf = 4.8π rad / 2.0 s = 9.42 rad/sThe closest answer choice is A) 9.5 rad/s.

Question 12

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

The answer of A wheel rotates about a fixed axis...

Question 13

A particle located at the position vector  m has a force  N acting on it. The torque about the origin is:
A)(1  )N.m
B)(5  )N.m
C)(-1  )N.m
D)(-5  )N.m
E)(2  + 3
 )N.m

Accepted Answer

The answer of A particle located at the position vector...

Question 14

A mass m = 4.0 kg is connected, as shown, by a light cord to a mass M = 6.0 kg, which slides on a smooth horizontal surface. The pulley rotates about a frictionless axle and has a radius R = 0.12 m and a moment of inertia I = 0.090 kg.m². The cord does not slip on the pulley. What is the magnitude of the acceleration of m? A)2.4 m/s2 B)2.8 m/s2 C)3.2 m/s2 D)4.2 m/s2 E)1.7 m/s2

Accepted Answer

The answer of A mass m = 4.0 kg is...

Question 15

A wheel (radius = 0.20 m) is mounted on a frictionless, horizontal axis. A light cord wrapped around the wheel supports a 0.50-kg object, as shown in the figure. When released from rest the object falls with a downward acceleration of 5.0 m/s2. What is the moment of inertia of the wheel? A)0.023 kg.m² B)0.027 kg.m² C)0.016 kg.m² D)0.019 kg.m² E)0.032 kg.m²

Accepted Answer

The answer of A wheel (radius = 0.20 m) is...

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 centre 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 centre of mass is given by:I = m1*r1^2 + m2*r2^2where m1 and m2 are the masses of the particles, r1 and r2 are the distances from the axis to the particles, and I is the moment of inertia.In this case, m1 = 0.20 kg, m2 = 0.30 kg, r1 = 1.0 m, and r2 = 1.0 m.Therefore, I = (0.20 kg)*(1.0 m)^2 + (0.30 kg)*(1.0 m)^2 = 0.50 kg.m^2.

Question 17

A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval?
A)2.5 s
B)4.0 s
C)3.5 s
D)3.0 s
E)4.5 s

Accepted Answer

Using the equation θ = ω₀t + 0.5αt², where θ = 80 rad, ω₀ = 0, α = 4.0 rad/s², and t = 4 s, we find the initial time t₀ by solving 80 = 0.5 * 4 * (t₀ + 4)² - 0.5 * 4 * t₀², which gives t₀ = 3.0 s.

Question 18

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 centre 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 =...

Question 19

Particles (mass of each = 0.20 kg) are placed at the 40-cm and 100-cm marks of a metre stick of negligible mass. This rigid body is free to rotate about a frictionless pivot at the 0-cm end. The body is released from rest in the horizontal position. What is the initial angular acceleration of the body?
A)12 rad/s2
B)5.9 rad/s2
C)8.4 rad/s2
D)5.4 rad/s2
E)17 rad/s2

Accepted Answer

Explanation:The torque about the pivot is:$$	au = (0.20 	ext{ kg})(9.8 	ext{ m/s}^2)(0.40 	ext{ m}) + (0.20 	ext{ kg})(9.8 	ext{ m/s}^2)(1.00 	ext{ m}) = 27.44 	ext{ N$\cdot$m}$$The moment of inertia of the system about the pivot is:$$I = (0.20 	ext{ kg})(0.40 	ext{ m})^2 + (0.20 	ext{ kg})(1.00 	ext{ m})^2 = 0.28 	ext{ kg$\cdot$m}^2$$Therefore, the initial angular acceleration is:$$\alpha = \frac{	au}{I} = \frac{27.44 	ext{ N$\cdot$m}}{0.28 	ext{ kg$\cdot$m}^2} = 97.96 	ext{ rad/s}^2$$

Question 20

A uniform rod (mass = 2.0 kg, length = 0.60 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60$\degree$ below the horizontal?
A)15 rad/s2
B)12 rad/s2
C)18 rad/s2
D)29 rad/s2
E)23 rad/s2

Accepted Answer

The torque due to gravity when the rod is 60° below the horizontal is τ = mg(L/2)sin(60°). The moment of inertia of the rod about the pivot is I = (1/3)mL². Using τ = Iα, we find α = (3g/2L)sin(60°). Substituting the given values, α ≈ 12 rad/s².

Quiz 9: Rotational Motion

The disk in a DVD player has an angular velocity of 8.0 rad/s when it is turned off. The disk comes to rest 2.5 s after being turned off. Through how many radians does the disk rotate after being turned off? Assume constant angular acceleration.

A wheel starts from rest and rotates with a constant angular acceleration about a fixed axis. It completes the first revolution 6.0 s after it started. How long after it started will the wheel complete the second revolution?

A wheel (radius = 0.20 m) starts from rest and rotates with a constant angular acceleration of 2.0 rad/s2. At the instant when the angular velocity is equal to 1.2 rad/s, what is the magnitude of the total linear acceleration of a point on the rim of the wheel?

Two vectors lying in the xz plane are given by the equations and . The value of is:

Two vectors lying in the xy plane are given by the equations and . The value of is:

A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s2 turns through 2.4 revolutions during a 2.0-s time interval. What is the angular velocity at the end of this time interval?

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 particle located at the position vector m has a force N acting on it. The torque about the origin is:

A wheel (radius = 0.20 m) is mounted on a frictionless, horizontal axis. A light cord wrapped around the wheel supports a 0.50-kg object, as shown in the figure. When released from rest the object falls with a downward acceleration of 5.0 m/s2. What is the moment of inertia of the wheel?

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 centre of mass?

A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval?

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 centre of mass? Treat the mass as particles.

A uniform rod (mass = 2.0 kg, length = 0.60 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60 $\degree$ below the horizontal?