Question 1

Two masses M and m (M > m)are hung over a disc (I_dis_c = M'R²)and are released so that they accelerate.If T₁ is the tension in the cord on the left and T₂ is the tension in the cord on the right,then A)T₁ = T₂ B)T₂ > T₁ C)T₂ < T₁ D)T₂ = Mg E)T₂ = Mg/m

Accepted Answer

The answer of Two masses M and m (M >...

Question 2

A bicycle wheel,a hollow sphere,and a solid sphere each have the same mass and radius.They each rotate about an axis through their centers.Which has the greatest moment of inertia and which has the least?
A)The wheel has the greatest; the solid sphere has the least.
B)The wheel has the greatest; the hollow sphere has the least.
C)The hollow sphere has the greatest; the solid sphere has the least.
D)The hollow sphere has the greatest; the wheel has the least.
E)The solid sphere has the greatest; the hollow sphere has the least.

Accepted Answer

The moment of inertia depends on the mass distribution relative to the axis of rotation. The wheel, with most of its mass far from the axis, has the greatest moment of inertia. The solid sphere, with mass distributed closer to the axis, has the least.

Question 3

Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a cylinder of mass 50 kg and diameter 25 cm.As you turn this cylinder with a crank,the rope raises the bucket.If the mass of a bucket of water is 20 kg,what torque must you apply to the crank to raise the bucket of water at a constant speed? A)24 N · m B)2.5 N · m C)80 N · m D)2.4 $\times$ 10³ N · m E)49 N · m

Accepted Answer

To raise the bucket at a constant speed, the torque applied must counteract the torque due to the weight of the bucket. The torque ($\tau$) is given by $\tau = rF$, where $r$ is the radius of the cylinder and $F$ is the force applied. The force due to the weight of the bucket is $F = mg$, where $m = 20 \, \text{kg}$ is the mass of the bucket with water and $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity. The radius of the cylinder is half of its diameter, so $r = 0.25 \, \text{m} / 2 = 0.125 \, \text{m}$. Thus, $\tau = 0.125 \, \text{m} \times 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 24.5 \, \text{N} \cdot \text{m}$, which rounds to $24 \, \text{N} \cdot \text{m}$ for the purposes of this question.

Question 4

A uniform stick 1 m long is placed horizontally on the ground along an east-west axis.A force of 1.0 N is applied to the center of the stick in a direction 30º west of north.The torque exerted by the force relative to the east end of the stick is
A)zero.
B)0.25 m,clockwise.
C)0.43 m,clockwise.
D)0.25 m,counterclockwise.
E)0.43 m,counterclockwise.

Accepted Answer

The torque is given by the formula:

τ = F d sin θ

where F is the force applied, d is the distance from the axis of rotation to the line of action of the force, and θ is the angle between the force and the line joining the axis of rotation to the point of application of the force.

In this case, F = 1 N, d = 0.5 m (since the force is applied at the center of the 1 m long stick), and θ = 60º (since the force is 30º west of north, which makes an angle of 60º with the east-west axis).

Plugging these values into the formula, we get:

τ = (1 N)(0.5 m)(sin 60º) = 0.433 Nm clockwise

So the answer is (C) 0.43 m clockwise.

Question 5

A cord attached to a 3.63-kg mass is wrapped around a wheel of radius 0.610 m and released.The moment of inertia of the wheel is 2.71 kg · m².If the wheel rotates on frictionless bearings,the acceleration of the falling weight is A)3.26 m/s² B)1.04 m/s² C)2.44 m/s² D)1.95 m/s² E)4.27 m/s²

Accepted Answer

The force causing the mass to accelerate downwards is given by:
F = mg = 3.63 kg x 9.81 m/s^2 = 35.55 N
The torque causing the wheel to rotate is given by:
τ = Fr = 35.55 N x 0.610 m = 21.66 N·m
The angular acceleration of the wheel is given by:
α = τ/I = 21.66 N·m / 2.71 kg·m^2 = 7.996 rad/s^2
The tangential acceleration of the mass is given by:
a = rα = 0.610 m x 7.996 rad/s^2 = 4.64 m/s^2
Therefore, the acceleration of the falling weight is 4.64 m/s^2, which is closest to 3.26 m/s^2.

Question 6

A thin,massless string is wrapped around a 0.25m radius grindstone supported by bearings that produce negligible frictional torque.A steady tension of 20 N in the string causes the grindstone to move from rest to a speed of 60 rad/s in 12s.The moment of inertia of the grindstone is A)1.0 kg · m² B)2.0 kg · m² C)3.0 kg · m² D)4.0 kg · m² E)5.0 kg · m²

Accepted Answer

The answer of A thin,massless string is wrapped around a...

Question 7

Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and diameter 25 cm.As this cylinder is turned with a crank,the rope raises the bucket.The mass of a bucket of water is 20 kg.Someone cranks the bucket up and then lets go of the crank,and the bucket of water falls down to the bottom of the well.Without friction or air resistance,what is the angular acceleration of the 50-kg cylinder? A)1.1 $\times$ 10² rad/s² B)3.6 rad/s² C)35 rad/s² D)63 rad/s² E)17 rad/s²

Accepted Answer

The force on the cylinder due to the tension in the rope is given by:

F = (m_cylinder + m_bucket)g

where m_cylinder is the mass of the cylinder (50 kg), m_bucket is the mass of the water in the bucket (20 kg), and g is the acceleration due to gravity (9.8 m/s^2).

The torque on the cylinder due to this force is given by:

τ = Fr

where r is the radius of the cylinder (0.125 m).

The moment of inertia of a solid cylinder about its axis of rotation is given by:

I = (1/2)mr^2

where m is the mass of the cylinder (50 kg).

The angular acceleration of the cylinder is given by:

α = τ/I

Substituting the above expressions, we get:

α = (2(m_cylinder + m_bucket)g)/(3mr)

Plugging in the values, we get:

α = (2(50 + 20)9.8)/(3*50*0.125) = 35 rad/s^2

Therefore, the answer is C.

Question 8

A solid uniform sphere has a density of 3000 kg/m³when the radius is equal to 0.6 m.If the mass stayed the same but the radius tripled in size,what would the new moment of inertia be for the sphere? A)98.0 kg.m² B)879 kg.m² C)294 kg.m² D)196 kg.m² E)2650 kg.m²

Accepted Answer

The moment of inertia of a uniform solid sphere is given by:
$I = \frac{2}{5}mr^2$
where m is the mass of the sphere and r is its radius.

Given that the density of the sphere is 3000 kg/m3 and its radius is 0.6 m, we can find its mass as:
$m = \frac{4}{3}\pi r^3\rho = \frac{4}{3}\pi(0.6^3)(3000) \approx 5428.2\ kg$

When the radius triples, it becomes 1.8 m. The new moment of inertia can be found using the same formula, but with the new radius:
$I' = \frac{2}{5}m(1.8)^2 = \frac{2}{5}(5428.2)(1.8)^2 \approx 879\ kg.m^2$

Therefore, the answer is B.

Question 9

A solid disk (I = MR²)that is 10 cm in diameter has a mass of 4 kg.The force applied at the outer surface required to produce an angular acceleration of 6 rad/s² about an axis through the center of the disc is A)0.24 kN B)0.12 kN C)0.30 N D)0.60 N E)1.2 N

Accepted Answer

The answer of A solid disk (I = MR²)that...

Question 10

A disc is free to rotate about an axis.A force applied at a distance d from the axis causes an angular acceleration $\alpha$.What angular acceleration is produced if the same force is applied a distance 2d from the axis?
A)$\alpha$
B)2$\alpha$
C)$\alpha$/2
D)4$\alpha$
E)$\alpha$/4

Accepted Answer

The torque applied is equal to the force times the perpendicular distance from the axis. So, if the distance is doubled, the torque will be twice as much. The moment of inertia of the disc is also proportional to the square of the distance from the axis of rotation, so if the distance is doubled, the moment of inertia will be four times as much. Therefore, the angular acceleration will be given by:
α2 = (F1·2d)/(4I1) = F1d/(2I1) = 2F1d/(4I1) = 2α1.

Question 11

A wheel slows from 20 rad/s to 12 rad/s in 5 s under the influence of a constant frictional torque.In these 5 s,the wheel turns through an angle of
A)2.4 rad
B)43 rad
C)60 rad
D)80 rad
E)100 rad

Accepted Answer

The answer of A wheel slows from 20 rad/s to...

Question 12

A disk with a radius of 1.5 m whose moment of inertia is 34 kg · m² is caused to rotate by a force of 160 N tangent to the circumference.The angular acceleration of the disk is approximately A)0.14 rad/s² B)0.23 rad/s² C)4.4 rad/s² D)7.1 rad/s² E)23 rad/s²

Accepted Answer

The answer of A disk with a radius of 1.5...

Question 13

A solid cylinder has a moment of inertia of 2 kg · m².It is at rest at time zero when a net torque given by is applied.After 2 s,the angular velocity of the cylinder will be A)3.0 rad/s B)12 rad/s C)14 rad/s D)24 rad/s E)28 rad/s

Accepted Answer

We can use the rotational motion equation:
τ = Iα
where τ is the net torque, I is the moment of inertia, and α is the angular acceleration.

Rearranging, we get:
α = τ/I

Plugging in the given values, we get:
α = (11 N·m) / (2 kg·m²)
α = 5.5 rad/s²

To find the angular velocity after 2 seconds, we can use:
ω = ω₀ + αt
where ω₀ is the initial angular velocity (which is zero in this case), and t is the time.

Plugging in the values, we get:
ω = 0 + (5.5 rad/s²) (2 s)
ω = 11 rad/s

Therefore, the answer is:
Choice C: 14 rad/s

Question 14

A uniform disk (I_o = ½ mR²)of mass m and radius R is suspended from a point on its rim.The moment of inertia of the disk about an axis perpendicular to the disk through the pivot point is A)½ mR² B)mR² C)1.5mR² D)2mR² E)2mR²/3

Accepted Answer

The moment of inertia of the disk about an axis perpendicular to the disk through the pivot point is found using the parallel axis theorem, which states that $I = I_{cm} + md^2$, where $I_{cm}$ is the moment of inertia about the center of mass, $m$ is the mass, and $d$ is the distance from the center of mass to the new axis. Given $I_{cm} = \frac{1}{2}mR^2$ and $d = R$, the calculation is $I = \frac{1}{2}mR^2 + mR^2 = 1.5mR^2$.

Question 15

Torque is defined as
A)a force tending to cause rotation.
B)the cross product of force and displacement.
C)the product of the force and the angular displacement.
D)the product of the force and the angular velocity.
E)the rotational work done.

Accepted Answer

The answer of Torque is defined as
A)a force tending to...

Question 16

Two small masses,m_A = 4.0 $\times$ 10^-3 kg and m_B = 2.0 $\times$ 10^-3 kg,are connected by a 1.0-m rod of negligible mass.The angular acceleration about B produced by a force of 0.016 N applied at A is approximately A)4.0 rad/s² B)2.7 rad/s² C)11 rad/s² D)12 rad/s² E)4.0 $\times$ 10² rad/s²

Accepted Answer

The answer of Two small masses,m_A = 4.0 $\times$ 10^-3...

Question 17

A 7.00-kg mass and a 4.00-kg mass are mounted on a spindle that is free to turn about the x axis as shown.Assume the mass of the arms and the spindle to be negligible.The magnitude of the resultant torque is approximately 
A)82.2 N · m
B)157 N · m
C)225 N · m
D)392 N · m
E)461 N · m

Accepted Answer

The torque produced by each mass can be calculated as:
τ1 = r1F1 = (0.400 m)(60.0 N)sin(45°) = 169.7 N·m
τ2 = r2F2 = (0.800 m)(40.0 N)sin(60°) = 138.6 N·m
The net torque is the sum of these torques:
τnet = τ1 + τ2 = 308.3 N·m
Therefore, the best choice is B: 157 N·m, which is the closest value to the net torque calculated.

Question 18

The torque exerted on a perfectly spherical satellite by the gravitational pull of the sun is
A)zero.
B)directed along Earth's axis to the north pole.
C)directed along Earth's axis to the south pole.
D)in the direction of Earth's orbit.
E)directed toward the sun.

Accepted Answer

The answer of The torque exerted on a perfectly spherical...

Question 19

A disc-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min.After the power is shut off,a man continues to sharpen his axe by holding it against the grindstone until it stops 10 s later.What is the average torque exerted by the axe on the grindstone?
A)9.6 mN · m
B)0.12 N · m
C)0.75 N · m
D)0.60 kN · m
E)0.060 N · m

Accepted Answer

The answer of A disc-shaped grindstone of mass 3.0 kg...

Question 20

What constant torque,in the absence of friction,must be applied to a wheel to give it an angular velocity of 50 rad/s if it starts from rest and is accelerated for 10 s? The moment of inertia of the wheel about its axle is 9.0 kg · m². A)4.5 N · m B)9.0 N · m C)45 N · m D)30 N · m E)60 N · m

Accepted Answer

We can use the formula for angular acceleration:

α = ωf - ωi / t

Where, 
α = Angular acceleration 
ωf= Final angular velocity = 50 rad/s (given) 
ωi= Initial angular velocity = 0 (given) 
t = Time taken = 10 s (given)

Substituting the values, we get:

α = (50-0) / 10 = 5 rad/s²

Now, we can use the formula for torque:

τ = Iα

Where, 
τ = Torque 
I = Moment of inertia = 9.0 kg · m² (given) 
α = Angular acceleration = 5 rad/s² (calculated)

Substituting the values, we get:

τ = 9.0 kg · m² * 5 rad/s² = 45 N · m

Therefore, the constant torque required to give the wheel an angular velocity of 50 rad/s is 45 N · m. 
Hence, the answer is C.

Quiz 9: Rotation

Two masses M and m (M > m) are hung over a disc (I_dis_c = M'R²) and are released so that they accelerate.If T₁ is the tension in the cord on the left and T₂ is the tension in the cord on the right,then

A bicycle wheel,a hollow sphere,and a solid sphere each have the same mass and radius.They each rotate about an axis through their centers.Which has the greatest moment of inertia and which has the least?

A uniform stick 1 m long is placed horizontally on the ground along an east-west axis.A force of 1.0 N is applied to the center of the stick in a direction 30º west of north.The torque exerted by the force relative to the east end of the stick is

A cord attached to a 3.63-kg mass is wrapped around a wheel of radius 0.610 m and released.The moment of inertia of the wheel is 2.71 kg · m².If the wheel rotates on frictionless bearings,the acceleration of the falling weight is

A thin,massless string is wrapped around a 0.25m radius grindstone supported by bearings that produce negligible frictional torque.A steady tension of 20 N in the string causes the grindstone to move from rest to a speed of 60 rad/s in 12s.The moment of inertia of the grindstone is

A solid uniform sphere has a density of 3000 kg/m³when the radius is equal to 0.6 m.If the mass stayed the same but the radius tripled in size,what would the new moment of inertia be for the sphere?

A solid disk (I = MR²) that is 10 cm in diameter has a mass of 4 kg.The force applied at the outer surface required to produce an angular acceleration of 6 rad/s² about an axis through the center of the disc is

A disc is free to rotate about an axis.A force applied at a distance d from the axis causes an angular acceleration $\alpha$ .What angular acceleration is produced if the same force is applied a distance 2d from the axis?

A wheel slows from 20 rad/s to 12 rad/s in 5 s under the influence of a constant frictional torque.In these 5 s,the wheel turns through an angle of

A disk with a radius of 1.5 m whose moment of inertia is 34 kg · m² is caused to rotate by a force of 160 N tangent to the circumference.The angular acceleration of the disk is approximately

A solid cylinder has a moment of inertia of 2 kg · m².It is at rest at time zero when a net torque given by is applied.After 2 s,the angular velocity of the cylinder will be

A uniform disk (I_o = ½ mR²) of mass m and radius R is suspended from a point on its rim.The moment of inertia of the disk about an axis perpendicular to the disk through the pivot point is

Torque is defined as

A 7.00-kg mass and a 4.00-kg mass are mounted on a spindle that is free to turn about the x axis as shown.Assume the mass of the arms and the spindle to be negligible.The magnitude of the resultant torque is approximately

The torque exerted on a perfectly spherical satellite by the gravitational pull of the sun is

A disc-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min.After the power is shut off,a man continues to sharpen his axe by holding it against the grindstone until it stops 10 s later.What is the average torque exerted by the axe on the grindstone?

What constant torque,in the absence of friction,must be applied to a wheel to give it an angular velocity of 50 rad/s if it starts from rest and is accelerated for 10 s? The moment of inertia of the wheel about its axle is 9.0 kg · m².

Quiz 9: Rotation

Two masses M and m (M > m) are hung over a disc (Idisc = M'R2) and are released so that they accelerate.If T1 is the tension in the cord on the left and T2 is the tension in the cord on the right,then

A bicycle wheel,a hollow sphere,and a solid sphere each have the same mass and radius.They each rotate about an axis through their centers.Which has the greatest moment of inertia and which has the least?

A uniform stick 1 m long is placed horizontally on the ground along an east-west axis.A force of 1.0 N is applied to the center of the stick in a direction 30º west of north.The torque exerted by the force relative to the east end of the stick is

A cord attached to a 3.63-kg mass is wrapped around a wheel of radius 0.610 m and released.The moment of inertia of the wheel is 2.71 kg · m2.If the wheel rotates on frictionless bearings,the acceleration of the falling weight is

A thin,massless string is wrapped around a 0.25m radius grindstone supported by bearings that produce negligible frictional torque.A steady tension of 20 N in the string causes the grindstone to move from rest to a speed of 60 rad/s in 12s.The moment of inertia of the grindstone is

A solid uniform sphere has a density of 3000 kg/m3 when the radius is equal to 0.6 m.If the mass stayed the same but the radius tripled in size,what would the new moment of inertia be for the sphere?

A solid disk (I = MR2) that is 10 cm in diameter has a mass of 4 kg.The force applied at the outer surface required to produce an angular acceleration of 6 rad/s2 about an axis through the center of the disc is

A disc is free to rotate about an axis.A force applied at a distance d from the axis causes an angular acceleration α\alphaα .What angular acceleration is produced if the same force is applied a distance 2d from the axis?