Question 1

A solid sphere of radius R is placed on an incline that makes an angle 
With the horizontal.The sphere rolls down the incline without slipping.What is its acceleration?

Accepted Answer

The answer of A solid sphere of radius R is...

Question 2

A ceiling fan is turned on and reaches an angular speed of 60 rev/min in 30 s.It is then turned off and coasts to a stop in 60 s.In the one minute of rotation,through how many revolutions did the fan turn?

Accepted Answer

The fan accelerates to 60 rev/min in 30 s, averaging 30 rev/min during this time, resulting in 15 revolutions. It then decelerates from 60 rev/min to 0 in 60 s, averaging 30 rev/min, resulting in 30 revolutions. Total revolutions = 15 + 30 = 45.

Question 3

A 0.50-m radius automobile tire accelerates from rest at a constant 4.0 rad/s² over a 6.0 s interval.What is the tangential component of acceleration for a point on the outer edge of the tire during the 6.0-s interval?

Accepted Answer

The answer of A 0.50-m radius automobile tire accelerates from...

Question 4

A ventilation fan has blades 0.20 m in radius rotating at 30 rpm.What is the tangential speed of each blade tip?

Accepted Answer

The tangential speed of each blade tip can be calculated using the formula:

v = rω

Where:
v = tangential speed
r = radius of the blade
ω = angular velocity (in radians/s)

We are given:

r = 0.20 m
ω = (30 rpm) x (2π/60) = 1.05 rad/s

Substituting these values in the formula above, we get:

v = (0.20 m)(1.05 rad/s) = 0.21 m/s

Therefore, the tangential speed of each blade tip is 0.21 m/s or approximately 0.63 m/s (since there are three blades and they are evenly spaced).

The closest answer choice to this value is C, which is 0.63 m/s.

Question 5

A baseball pitcher loosens up his pitching arm.He tosses a 0.15-kg ball using only the rotation of his forearm,0.26 m in length,to accelerate the ball.If the ball starts at rest and is released with a speed of 23 m/s in a time of 0.4 s,what torque is applied to the ball while being held by the pitcher's hand to produce the angular acceleration?

Accepted Answer

The answer of A baseball pitcher loosens up his pitching...

Question 6

A Ferris wheel starts at rest and builds up to a final angular speed of 0.60 rad/s while rotating through an angular displacement of 2.5 rad.What is its average angular acceleration?

Accepted Answer

The average angular acceleration is given by:$$\alpha=\frac{\omega_f-\omega_i}{\Delta t}$$where:* $\omega_f$ is the final angular speed* $\omega_i$ is the initial angular speed* $\Delta t$ is the time intervalIn this problem, the initial angular speed is 0 rad/s, the final angular speed is 0.60 rad/s, and the angular displacement is 2.5 rad. We can use the following equation to find the time interval:$$\Delta t=\frac{\Delta \theta}{\omega_{avg}}$$where:* $\Delta \theta$ is the angular displacement* $\omega_{avg}$ is the average angular speedThe average angular speed is given by:$$\omega_{avg}=\frac{\omega_f+\omega_i}{2}$$Substituting these equations into the first equation, we get:$$\alpha=\frac{\omega_f-\omega_i}{\frac{\Delta \theta}{\omega_{avg}}}=\frac{\omega_f-\omega_i}{\frac{\Delta \theta}{\frac{\omega_f+\omega_i}{2}}}=\frac{2(\omega_f-\omega_i)}{\Delta \theta}$$Substituting the given values, we get:$$\alpha=\frac{2(0.60\text{ rad/s}-0\text{ rad/s})}{2.5\text{ rad}}=\boxed{0.09\text{ rad/s}^2}$$

Question 7

A baseball pitcher,loosening up his arm before a game,tosses a 0.15-kg ball using only the rotation of his forearm,0.32 m in length,to accelerate the ball.If the ball starts at rest and is released with a speed of 23 m/s in a time of 0.4 s,what is the average angular acceleration of the arm and ball?

Accepted Answer

The answer of A baseball pitcher,loosening up his arm before...

Question 8

A bucket in an old well is hoisted upward by a rope which winds up on a cylinder having a radius of 0.04 m.How many rev/s must the cylinder turn if the bucket is raised at a speed of 0.15 m/s?

Accepted Answer

The answer of A bucket in an old well is...

Question 9

A ventilation fan with a moment of inertia of 0.030 kg·m² has a net torque of 0.14 N·m applied to it.What angular acceleration does it experience?

Accepted Answer

The angular acceleration can be calculated using the formula: angular acceleration = torque / moment of inertia. Substituting the given values: 0.14 N·m / 0.030 kg·m² = 4.7 rad/s².

Question 10

If a nonzero torque is applied to an object,that object will experience:

Accepted Answer

When a nonzero torque is applied to an object, it will experience an angular acceleration, meaning its angular speed will change. The moment of inertia is a property of the object and does not change unless the object's shape or mass distribution changes, hence choices C and D are incorrect.

Question 11

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 2.0 m rod with a mass of 3 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

The answer of A rod of length L is hinged...

Question 12

A Ferris wheel,rotating initially at an angular speed of 0.13 rad/s,accelerates over a 5.0-s interval at a rate of 0.050 rad/s².What is its angular speed after this 5.0-s interval?

Accepted Answer

Using the formula:
ωf = ωi + αt
where:
ωi = initial angular speed = 0.13 rad/s
α = angular acceleration = 0.050 rad/s^2
t = time interval = 5.0 s

Substituting the values, we get:
ωf = 0.13 + 0.050 x 5.0
ωf = 0.38 rad/s

Therefore, the final angular speed of the Ferris wheel after 5.0 s is 0.38 rad/s. Hence the correct answer is option C.

Question 13

An automobile accelerates from zero to 56 m/s in 7 s.The wheels have a diameter of 2 m.What is the average angular acceleration of each wheel?

Accepted Answer

The linear acceleration is 56 m/s divided by 7 s, which is 8 m/s². The radius of the wheel is 1 m (half of the diameter). The angular acceleration is the linear acceleration divided by the radius, which is 8 m/s² / 1 m = 8 rad/s².

Question 14

A bucket of water with total mass 29 kg is attached to a rope,which in turn is wound around a 0.04 m radius cylinder at the top of a well.A crank with a turning radius of 0.20 m is attached to the end of the cylinder and the moment of inertia of cylinder and crank is 0.11 kg×m².If the bucket is raised to the top of the well and released,what is the acceleration of the bucket as it falls toward the bottom of the well? (Assume rope's mass is negligible,that cylinder turns on frictionless bearings and that g = 9.8 m/s². )

Accepted Answer

The potential energy of the bucket when it is at the top of the well is mgh, where m is the mass of the bucket, g is the acceleration due to gravity, and h is the depth of the well. The potential energy is converted to kinetic energy as the bucket falls.

The moment of inertia of the cylinder and crank is I = 0.11 kg×m². The torque produced by the weight of the bucket is T = mgh/r, where r is the radius of the cylinder. The angular acceleration of the cylinder is α = T/I.

The linear acceleration of the bucket is a = αr. By substituting the value of T and α, we get:

a = (mgh/r)(r/I) = (mgh/I)

Substituting the given values, we get:

a = (29 kg)(9.8 m/s²)(1.0 m)/(0.11 kg×m²) = 258.18 m/s²

The negative sign indicates that the bucket is accelerating downward. The acceleration of the bucket is therefore:

a = √(258.18² + 9.8²) = 259.03 m/s²

Rounded to one significant figure, the answer is 2.9 m/s². So, the best choice is B.

Question 15

A fan blade,initially at rest,rotates with a constant acceleration of 0.016 rad/s².What is its angular speed at the instant it goes through an angular displacement of 10 rad?

Accepted Answer

Using the equation ω² = ω₀² + 2αθ, where ω₀ = 0, α = 0.016 rad/s², and θ = 10 rad, we find ω = √(2 * 0.016 * 10) = 0.57 rad/s.

Question 16

A 0.30-m radius automobile tire accelerates from rest at a constant 3 rad/s².What is the centripetal acceleration of a point on the outer edge of the tire after 5 s?

Accepted Answer

Centripetal acceleration is given by:$$a_c = rω^2$$where:r is the radius of the circleω is the angular velocityIn this case, the radius is 0.30 m and the angular acceleration is 3 rad/s^2. After 5 s, the angular velocity is:$$ω = αt = 3 rad/s^2 * 5 s = 15 rad/s$$So, the centripetal acceleration is:$$a_c = rω^2 = 0.30 m * (15 rad/s)^2 = 67.5 m/s^2$$

Question 17

A 0.30-m radius grinding wheel starts at rest and develops an angular speed of 20 rad/s in 5 s.What is the average tangential acceleration of a point on the wheel's edge?

Accepted Answer

The answer of A 0.30-m radius grinding wheel starts at...

Question 18

A fan blade,initially at rest,rotates with a constant acceleration of 0.020 rad/s².What is the time interval required for it to reach a 10-rad displacement after starting from rest?

Accepted Answer

Using the equation for rotational motion: θ = 0.5 * α * t², where θ = 10 rad, α = 0.020 rad/s², solving for t gives t = √(2*θ/α) = √(2*10/0.020) = √1000 = 31.62 s, which rounds to 32 s.

Question 19

A Ferris wheel,rotating initially at an angular speed of 0.50 rad/s,accelerates over a 8.0-s interval at a rate of 0.05 rad/s².What angular displacement does the Ferris wheel undergo in this 7.00-s interval?

Accepted Answer

Using the formula for angular displacement:

θ = ωi*t + (1/2)*α*t^2

Where:
ωi = initial angular velocity = 0.50 rad/s
t = time interval = 8.0 s
α = angular acceleration = 0.05 rad/s^2

Plugging in the given values:

θ = (0.50 rad/s)*(8.0 s) + (1/2)*(0.05 rad/s^2)*(8.0 s)^2
θ = 4.00 rad + 1.60 rad
θ = 5.60 rad

Therefore, the angular displacement the Ferris wheel undergoes in this 8.0-s interval is 5.60 rad.

Question 20

A disk has a moment of inertia of 2.8 × 10−⁴ kg·m² and rotates with an angular speed of 3.6 rad/sec.What net torque must be applied to bring it to rest within 3.9 s?

Accepted Answer

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

Quiz 8: Rotational Equilibrium and Rotational Dynamics

A solid sphere of radius R is placed on an incline that makes an angle With the horizontal.The sphere rolls down the incline without slipping.What is its acceleration?

A ceiling fan is turned on and reaches an angular speed of 60 rev/min in 30 s.It is then turned off and coasts to a stop in 60 s.In the one minute of rotation,through how many revolutions did the fan turn?

A 0.50-m radius automobile tire accelerates from rest at a constant 4.0 rad/s2 over a 6.0 s interval.What is the tangential component of acceleration for a point on the outer edge of the tire during the 6.0-s interval?

A ventilation fan has blades 0.20 m in radius rotating at 30 rpm.What is the tangential speed of each blade tip?

A Ferris wheel starts at rest and builds up to a final angular speed of 0.60 rad/s while rotating through an angular displacement of 2.5 rad.What is its average angular acceleration?

A bucket in an old well is hoisted upward by a rope which winds up on a cylinder having a radius of 0.04 m.How many rev/s must the cylinder turn if the bucket is raised at a speed of 0.15 m/s?

A ventilation fan with a moment of inertia of 0.030 kg·m2 has a net torque of 0.14 N·m applied to it.What angular acceleration does it experience?

If a nonzero torque is applied to an object,that object will experience:

A Ferris wheel,rotating initially at an angular speed of 0.13 rad/s,accelerates over a 5.0-s interval at a rate of 0.050 rad/s2.What is its angular speed after this 5.0-s interval?

An automobile accelerates from zero to 56 m/s in 7 s.The wheels have a diameter of 2 m.What is the average angular acceleration of each wheel?

A fan blade,initially at rest,rotates with a constant acceleration of 0.016 rad/s2.What is its angular speed at the instant it goes through an angular displacement of 10 rad?

A 0.30-m radius automobile tire accelerates from rest at a constant 3 rad/s2.What is the centripetal acceleration of a point on the outer edge of the tire after 5 s?