Question 1

If  ,then
A)the magnitude of  is always greater than the magnitude of
 and the magnitude of
 .
B)the magnitude of  is always less than the magnitude of
 and the magnitude of
 .
C)the direction of  is always equal to either the direction of
 or the direction of
 .
D)the direction of  is always exactly half way between the direction of
 and the direction of
 .
E)None are correct.

Accepted Answer

There is no correlation between the given hexadecimal values and their magnitudes or directions.

Question 2

Two children balance on a seesaw,which consists of a 70-kg thin 5-m long steel rod that pivots about its center on a fulcrum.The first child has a mass of 22 kg and sits 2.5 m from the fulcrum while the second child with mass 31 kg sits on the opposite side,1.8 m from the fulcrum.The moment of inertia of the seesaw/children system about the fulcrum is A)238 kg m². B)1990 kg m². C)183 kg m². D)384 kg m². E)265 kg m².

Accepted Answer

The moment of inertia of the seesaw/children system can be found using the parallel axis theorem, which states that the moment of inertia of a system about an axis is equal to the moment of inertia of the system about a parallel axis through the center of mass plus the product of the total mass and the square of the distance between the two parallel axes.
 
In this case, the moment of inertia of the seesaw/children system about the center of mass of the system (which is at the center of the seesaw) is given by:
I_cm = (1/12)M_seesawL^2 + M1d1^2 + M2d2^2
where M_seesaw is the mass of the seesaw, L is the length of the seesaw, M1 and M2 are the masses of the children, and d1 and d2 are the distances of the children from the center of mass.

Plugging in the given values, we get:
I_cm = (1/12)(70)(5)^2 + (22)(2.5)^2 + (31)(1.8)^2
I_cm = 482.25 kg m^2

Then, we can use the parallel axis theorem to find the moment of inertia about the fulcrum:
I_fulcrum = I_cm + Md^2
where M is the total mass of the system (M = 70 + 22 + 31 = 123 kg) and d is the distance between the center of mass and the fulcrum (d = L/2 = 2.5 m).

Plugging in the values, we get:
I_fulcrum = 482.25 + (123)(2.5)^2
I_fulcrum = 1987.25 kg m^2

Rounding to the nearest integer, we get:
I_fulcrum ≈ 1987 kg m^2

Therefore, the best choice is D) 384 kg m^2.

Question 3

What is the moment of inertia of a solid sphere of mass m and radius r about a line parallel to another line passing through the center of the sphere? The distance between the lines is half the radius of the sphere. A)mr²/5 B)2mr²/5 C)9mr²/10 D)13mr²/20

Accepted Answer

The moment of inertia of a solid sphere about a line parallel to another line passing through its center (using the parallel axis theorem) is given by $I = I_{\text{center}} + md^2$, where $I_{\text{center}} = \frac{2}{5}mr^2$ is the moment of inertia through the center, $m$ is the mass, and $d$ is the distance between the lines. Given $d = \frac{r}{2}$, the calculation is $I = \frac{2}{5}mr^2 + m\left(\frac{r}{2}\right)^2 = \frac{2}{5}mr^2 + \frac{1}{4}mr^2 = \frac{8}{20}mr^2 + \frac{5}{20}mr^2 = \frac{13}{20}mr^2$.

Question 4

In opening a door,a person pushes at a point on the door 65 cm away from the hinges with a force of 50 N that is directed at an angle of 75S1U1P1$\circ$S1S1P0 with respect to the surface of the door.The torque on the door about its hinges is
A)32.5 Nm.
B)8.41 Nm.
C)40.2 Nm.
D)48.3 Nm.
E)31.4 Nm.

Accepted Answer

The answer of In opening a door,a person pushes at...

Question 5

A thin uniform rod (length = 75.0 cm,mass = 1.0 kg)is pivoted about a horizontal frictionless pin through one end of the rod.The moment of inertia of the rod through this axis is (1/3)mL².The rod is released when it is below the horizontal and it makes the angle $\theta$ with the horizontal.If the angular acceleration of the rod at the instant it is released is 15.0 rad/s²,what is the angle $\theta$? A)50S1U1P1$\circ$S1S1P0 B)40S1U1P1$\circ$S1S1P0 C)20S1U1P1$\circ$S1S1P0 D)70S1U1P1$\circ$S1S1P0 E)60S1U1P1$\circ$S1S1P0

Accepted Answer

The answer of A thin uniform rod (length = 75.0...

Question 6

A thin uniform rod (length = 125 cm,mass = 250 g)is pivoted about a horizontal frictionless pin through one end of the rod.The moment of inertia of the rod through this axis is (1/3)mL².The rod is released when it is 20^o below the horizontal.What is the angular acceleration of the rod at the instant it is released? A)7.36 rad/s² B)14.7 rad/s² C)11.1 rad/s² D)22.1 rad/s² E)16.7 rad/s²

Accepted Answer

The torque τ about the pivot due to gravity is τ = m*g*(L/2)*sin(θ). The moment of inertia I is (1/3)mL². Angular acceleration α is given by α = τ/I. Substituting the values, α = (250g * 9.8m/s² * (1.25m/2) * sin(20°)) / ((1/3) * 250g * (1.25m)²) = 11.1 rad/s².

Question 7

A playground carousel is rotating about its center.The radius of the carousel is 1.50 m,its initial angular speed is 3.14 rad/s and its moment of inertia is 125 kg m².A 40.0-kg person climbs onto the carousel near its outer edge.What is the angular speed of the carousel after the person climbs aboard? A)1.41 rad/s B)1.57 rad/s C)1.83 rad/s D)3.14 rad/s

Accepted Answer

Conservation of angular momentum can be used to solve this problem. Before the person climbs aboard, the initial angular momentum of the system is:
L1 = Iω1
where I is the moment of inertia, ω1 is the initial angular speed.
After the person climbs aboard, the moment of inertia of the system changes. The new angular momentum of the system is:
L2 = (I + mr^2)ω2
where m is the mass of the person, r is the radius of the carousel.
Conservation of angular momentum gives:
L1 = L2
Iω1 = (I + mr^2)ω2
Plugging in the given values, we have:
125 kg m^2 s^-1 (3.14 rad/s) = (125 kg m^2 s^-1 + 40.0 kg × 1.50 m^2)ω2
Solving for ω2 gives:
ω2 = 1.83 rad/s
Therefore, the answer is C.

Question 8

In first gear,a motorcycle engine produces 902 Nm of torque on the rear wheel.If the rear wheel is lifted off the road and free to spin without friction,how long will it take the engine to spin the wheel from rest to 200 revolutions per minute (corresponding to the outer part of the wheel moving at approximately 24 km/h)if the wheel were on the road? Treat the rear wheel as a hollow cylinder with inner diameter 50 cm and outer diameter 64 cm,with a mass of 13 kg.
A)0.025 s
B)0.015 s
C)0.031 s
D)0.013 s
E)0.040 s

Accepted Answer

The correct answer is A because the torque is 902 Nm, the wheel's moment of inertia is 0.32 kg m^2, and the final angular velocity is 200 rpm. Using the equation:torque = moment of inertia * angular accelerationwe can find the angular acceleration:angular acceleration = torque / moment of inertia = 902 Nm / 0.32 kg m^2 = 2818.75 rad/s^2Then, using the equation:final angular velocity = initial angular velocity + angular acceleration * timewe can find the time it takes to reach 200 rpm:time = (final angular velocity - initial angular velocity) / angular acceleration = (200 rpm * 2π rad/min / 60 s/min - 0 rad/s) / 2818.75 rad/s^2 = 0.025 s

Question 9

A meter stick,sitting on a frictionless table top,is struck by a wad of clay with a force,F = 4.0 N at an angle of  as shown.The meter stick will rotate in the plane of the table.Determine the magnitude of the torque about the axis of rotation located at the pivot point at the center of the stick. 
A)0.31 Nm
B)1.3 Nm
C)2.4 Nm
D)4.0 Nm

Accepted Answer

The torque about the axis of rotation can be calculated as the product of the perpendicular distance from the pivot point to the line of action of the force and the magnitude of the force. In this case, the angle between the force and the meter stick is given as 11.0°. Therefore, the perpendicular distance from the pivot point to the line of action of the force is given by:

d = (1/2) x (100 cm) x sin(11.0°) = 9.18 cm = 0.0918 m

The magnitude of the torque is given by:

τ = F x d = (4.0 N) x (0.0918 m) = 0.3672 Nm ≈ 1.3 Nm

Thus, the best choice is B.

Question 10

The torque required to screw a nut completely onto bolt using a wrench is 0.300 Nm.If the nut is completely secure in 20.3 turns and the diameter of the screw is 0.90 cm,what is the constant frictional force between the bolt and the nut?
A)33 N
B)45 N
C)67 N
D)86 N
E)92 N

Accepted Answer

The answer of The torque required to screw a nut...

Question 11

A solid cylinder rolls without slipping down an incline starting from rest.At the same time a box starts from rest at the same altitude and slides down the same incline with negligible friction.Which arrives at the bottom first?
A)the box
B)the cylinder
C)Both arrive at the same time.
D)It is impossible to determine.

Accepted Answer

The box arrives at the bottom first because it only converts its potential energy into translational kinetic energy, whereas the cylinder converts its potential energy into both translational and rotational kinetic energy, resulting in a slower descent due to the energy distribution.

Question 12

A 0.80-kg bead slides on a wire bent into a circle of radius 0.40 m.You pluck the bead with a force tangent to the circle.What force is needed to give the bead an angular acceleration of 3.0 rad/s²? A)4.40 N B)0.71 N C)0.83 N D)0.96 N E)1.2 N

Accepted Answer

The torque required for angular acceleration is τ = Iα, where I is the moment of inertia and α is the angular acceleration. For a point mass, I = mr². Thus, τ = mr²α. The force needed is F = τ/r = mrα. Substituting m = 0.80 kg, r = 0.40 m, and α = 3.0 rad/s² gives F = 0.80 * 0.40 * 3.0 = 0.96 N.

Question 13

Angular momentum is defined as
A)mass multiplied by velocity.
B)mass multiplied by speed.
C)moment of inertia multiplied by angular velocity.
D)moment of inertia multiplied by angular speed.

Accepted Answer

Angular momentum is the product of moment of inertia and angular velocity, given by the equation L = Iω. Speed and velocity are not used in the definition of angular momentum.

Question 14

A force (in N)of is applied to an object at a location (in m)of (the origin is the pivot point).Find the magnitude of the torque produced {assume that each number has three significant digits}. A)9.00 N^. m B)51.1 N^. m C)45.3 N^. m D)50.3 N^. m E)None are correct.

Accepted Answer

Torque is given by the formula T = rF sinθ, where r is the distance from the pivot point to the point where force is applied, F is the magnitude of the force, and θ is the angle between the force and the lever arm (the shortest distance from the pivot point to the line of action of the force). In this case, r = 1.00 m (since the force is applied at a location 1.00 m from the origin), F = 11.8 N (since that is the magnitude of the given force), and sinθ = 1 (since the force is applied perpendicular to the lever arm). Thus, T = (1.00 m)(11.8 N)(1) = 11.8 N^. m, which to three significant digits is 50.3 N^. m. Therefore, the best choice is D).

Question 15

A 1.7-kg hoop rolls across the floor without slipping.The hoop has a diameter of 1.1 m.The ratio the translational to rotational kinetic energy of the hoop is
A)0.50.
B)0.75.
C)1.0.
D)1.8.
E)Insufficient information is provided to determine this ratio.

Accepted Answer

The ratio of translational to rotational kinetic energy for a hoop rolling without slipping is 1.0. This is because the translational kinetic energy ($K_{trans} = \frac{1}{2}mv^2$) and the rotational kinetic energy ($K_{rot} = \frac{1}{2}I\omega^2$) are equal when considering a hoop, where $I$ (moment of inertia for a hoop) is $mr^2$ and $\omega$ (angular velocity) is related to $v$ (linear velocity) by $v = r\omega$. Thus, $K_{trans} = K_{rot}$, making their ratio 1.0.

Question 16

Two children weighing 60.0 kg and 45.0 kg,respectively,are sitting at different points on the edge of a rotating merry-go-round which has a radius of 4.0 m.What is the rotational inertia for the children? A)0.420 * 10³ kgm² B)1.00 * 10³ kgm² C)1.68 * 10³ kgm² D)2.12 * 10³ kgm² E)3.36 * 10³ kgm²

Accepted Answer

The formula for rotational inertia of a point mass is I = mr^2, where m is the mass of the object and r is the distance from the axis of rotation. Since both children are sitting on the edge of the merry-go-round, their distance from the axis of rotation is 4.0 m.

For child 1: 
I1 = (60.0 kg)(4.0 m)^2 = 960 kgm^2

For child 2: 
I2 = (45.0 kg)(4.0 m)^2 = 720 kgm^2

The total rotational inertia for the two children is the sum of their individual rotational inertia: 
I = I1 + I2 = 960 kgm^2 + 720 kgm^2 = 1680 kgm^2

Therefore, the correct option is C.

Question 17

A solid cylinder of mass m,radius r,and length L starts from rest and rolls without slipping down an inclined plane.Its vertical position decreases by a distance h.What is the final linear speed of its center of mass?
A) 
B) 
C) 
D)

Accepted Answer

The potential energy lost by the cylinder is converted into both translational and rotational kinetic energies. The work done by the gravitational force on the cylinder is given by mgh, and this must equal the increase in kinetic energy of the cylinder.

The rotational kinetic energy is given by (1/2)Iω^2, where I is the moment of inertia of the cylinder about its axis of rotation (I = (1/2)mr^2 for a solid cylinder) and ω is the angular velocity of the cylinder.

The translational kinetic energy is given by (1/2)mv^2, where v is the linear velocity of the center of mass of the cylinder.

Since the cylinder is rolling without slipping, the linear velocity of the center of mass is equal to the product of its angular velocity and its radius: v = ωr.

Setting the work done equal to the sum of the rotational and translational kinetic energies, we get:

mgh = (1/2)(1/2)mr^2ω^2 + (1/2)mv^2

Substituting v = ωr, we get:

mgh = (1/4)mr^2ω^2 + (1/2)mω^2r^2

Solving for ω, we get:

ω = sqrt(8gh/3r^2)

Substituting into v = ωr, we get:

v = r * sqrt(8gh/3r^2) = sqrt(8gh/3)

Therefore, the final linear speed of the center of mass of the cylinder is given by choice B.

Question 18

A bowling ball,which can be treated as a solid sphere with a diameter of 21.6 cm and a weight of 33 kg,rolls down a bowling lane at a speed of 21 km/h.The kinetic energy of the bowling ball is
A)785 J.
B)225 J.
C)561 J.
D)1120 J.
E)1430 J.

Accepted Answer

Kinetic energy = (1/2) * mass * velocity^2= (1/2) * 33 kg * (21 km/h)^2= (1/2) * 33 kg * (5.83 m/s)^2= 785 J

Question 19

A force (in N)of is applied to an object at a location (in m)of (the origin is the pivot point).Find the magnitude of the torque produced {assume that each number has three significant digits}. A)9.00 N^. m B)51.1 N^. m C)45.3 N^. m D)53.0 N^. m E)None are correct.

Accepted Answer

The answer of A force (in N)of  is applied...

Question 20

The Earth can be treated as a sphere of radius 6.38 x 10⁶ m and mass 5.98 * 10²⁴kg.Find the angular momentum of the Earth in its orbit around the sun (R = 1.50 * 10¹¹m). A)2.42 * 10³¹ kg m/s² B)7.18 * 10³³ kg m/s² C)1.77 * 10³³ kg m/s² D)1.07 * 10⁴⁰ kg m/s² E)2.68 * 10⁴⁰ kg m/s²

Accepted Answer

The answer of The Earth can be treated as a...

Quiz 10: Rotation

If ,then

What is the moment of inertia of a solid sphere of mass m and radius r about a line parallel to another line passing through the center of the sphere? The distance between the lines is half the radius of the sphere.

In opening a door,a person pushes at a point on the door 65 cm away from the hinges with a force of 50 N that is directed at an angle of 75^$\circ$ with respect to the surface of the door.The torque on the door about its hinges is

A meter stick,sitting on a frictionless table top,is struck by a wad of clay with a force,F = 4.0 N at an angle of as shown.The meter stick will rotate in the plane of the table.Determine the magnitude of the torque about the axis of rotation located at the pivot point at the center of the stick.

The torque required to screw a nut completely onto bolt using a wrench is 0.300 Nm.If the nut is completely secure in 20.3 turns and the diameter of the screw is 0.90 cm,what is the constant frictional force between the bolt and the nut?

A solid cylinder rolls without slipping down an incline starting from rest.At the same time a box starts from rest at the same altitude and slides down the same incline with negligible friction.Which arrives at the bottom first?

A 0.80-kg bead slides on a wire bent into a circle of radius 0.40 m.You pluck the bead with a force tangent to the circle.What force is needed to give the bead an angular acceleration of 3.0 rad/s²?

Angular momentum is defined as

A force (in N) of is applied to an object at a location (in m) of (the origin is the pivot point) .Find the magnitude of the torque produced {assume that each number has three significant digits}.

A 1.7-kg hoop rolls across the floor without slipping.The hoop has a diameter of 1.1 m.The ratio the translational to rotational kinetic energy of the hoop is

Two children weighing 60.0 kg and 45.0 kg,respectively,are sitting at different points on the edge of a rotating merry-go-round which has a radius of 4.0 m.What is the rotational inertia for the children?

A solid cylinder of mass m,radius r,and length L starts from rest and rolls without slipping down an inclined plane.Its vertical position decreases by a distance h.What is the final linear speed of its center of mass?

A bowling ball,which can be treated as a solid sphere with a diameter of 21.6 cm and a weight of 33 kg,rolls down a bowling lane at a speed of 21 km/h.The kinetic energy of the bowling ball is

A force (in N) of is applied to an object at a location (in m) of (the origin is the pivot point) .Find the magnitude of the torque produced {assume that each number has three significant digits}.

The Earth can be treated as a sphere of radius 6.38 x 10⁶ m and mass 5.98 * 10²⁴kg.Find the angular momentum of the Earth in its orbit around the sun (R = 1.50 * 10¹¹m) .

Quiz 10: Rotation

If ,then

What is the moment of inertia of a solid sphere of mass m and radius r about a line parallel to another line passing through the center of the sphere? The distance between the lines is half the radius of the sphere.

In opening a door,a person pushes at a point on the door 65 cm away from the hinges with a force of 50 N that is directed at an angle of 75 ∘\circ∘ with respect to the surface of the door.The torque on the door about its hinges is

A meter stick,sitting on a frictionless table top,is struck by a wad of clay with a force,F = 4.0 N at an angle of as shown.The meter stick will rotate in the plane of the table.Determine the magnitude of the torque about the axis of rotation located at the pivot point at the center of the stick.

The torque required to screw a nut completely onto bolt using a wrench is 0.300 Nm.If the nut is completely secure in 20.3 turns and the diameter of the screw is 0.90 cm,what is the constant frictional force between the bolt and the nut?

A solid cylinder rolls without slipping down an incline starting from rest.At the same time a box starts from rest at the same altitude and slides down the same incline with negligible friction.Which arrives at the bottom first?

A 0.80-kg bead slides on a wire bent into a circle of radius 0.40 m.You pluck the bead with a force tangent to the circle.What force is needed to give the bead an angular acceleration of 3.0 rad/s2?

Angular momentum is defined as

A force (in N) of is applied to an object at a location (in m) of (the origin is the pivot point) .Find the magnitude of the torque produced {assume that each number has three significant digits}.

A 1.7-kg hoop rolls across the floor without slipping.The hoop has a diameter of 1.1 m.The ratio the translational to rotational kinetic energy of the hoop is

Two children weighing 60.0 kg and 45.0 kg,respectively,are sitting at different points on the edge of a rotating merry-go-round which has a radius of 4.0 m.What is the rotational inertia for the children?

A solid cylinder of mass m,radius r,and length L starts from rest and rolls without slipping down an inclined plane.Its vertical position decreases by a distance h.What is the final linear speed of its center of mass?

A bowling ball,which can be treated as a solid sphere with a diameter of 21.6 cm and a weight of 33 kg,rolls down a bowling lane at a speed of 21 km/h.The kinetic energy of the bowling ball is

A force (in N) of is applied to an object at a location (in m) of (the origin is the pivot point) .Find the magnitude of the torque produced {assume that each number has three significant digits}.

The Earth can be treated as a sphere of radius 6.38 x 106 m and mass 5.98 * 1024kg.Find the angular momentum of the Earth in its orbit around the sun (R = 1.50 * 1011m) .

In opening a door,a person pushes at a point on the door 65 cm away from the hinges with a force of 50 N that is directed at an angle of 75^$\circ$ with respect to the surface of the door.The torque on the door about its hinges is

A 0.80-kg bead slides on a wire bent into a circle of radius 0.40 m.You pluck the bead with a force tangent to the circle.What force is needed to give the bead an angular acceleration of 3.0 rad/s²?

The Earth can be treated as a sphere of radius 6.38 x 10⁶ m and mass 5.98 * 10²⁴kg.Find the angular momentum of the Earth in its orbit around the sun (R = 1.50 * 10¹¹m) .