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ATHS Ch.08 ( Torque and Angular Momentum ) G11ES

8.1 Torque

Only the force component perpendicular to the line connecting the axis of rotation and the point of application of the force results in a torque about that axis. a. The lever arm is the perpendicular distance from the axis of rotation or revolution to the line of application of the force. b. The magnitude of the torque is the product of the magnitude of the lever arm and the magnitude of the force .

Torque The length pg the position vector Force applied

( The force arm )

(The distance between the axis of rotation and the force applied)

( N.m ) ( m ) ( N ) c. The net torque on a balanced system is zero .

Clockwise rotation ( - ) d. Torque vector quantity

Counterclockwise rotation ( + )

1


Q 1 : Two disgruntled businesspeople are trying to use a revolving door, as in Figure below. The woman on the left exerts a force of 625 N perpendicular to the door and

1.20 m from the hub’s center, while the man on the right exerts a force of 8.50 x10 2 N perpendicular to the door and 0.800 m from the hub’s center. Find the net torque on the revolving door.

Q2 : A businessman enters the same revolving door on the right, pushing with 576 N of force directed perpendicular to the door and 0.700 m from the hub, while a boy exerts a force of 365 N perpendicular to the door, 1.25 m to the left of the hub. Find

(a) the torques exerted by each person.

(b) the net torque on the door.

2


Q3 : A board of negligible mass supports a mass weighing 20 N at a distance of 2 m from the support at point A. How much weight must be placed at point B which is 5 m from point A in order for the rod to remain horizontal? e. If there is an angle ( Ѳ ) between F and r :

I .The magnitude can be calculated as : angle between F & r

II. The direction can be determined by the right-hand rule :

1. Point the fingers of your right hand in the direction of r .

2. Curl your fingers toward the direction of vector F .

3. Your thumb then points approximately in the direction of the torque, in this case out of the page.

3


Q4 :

The right-hand rule: Point the fingers of your right hand along r S and curl them in the direction of F S . Your thumb then points in the direction of the torque

(out of the page, in this case). Note that either u or u9 can be used in the definition of torque.

(a) A man applies a force of F = 3.00 x 10 2 N at a n angle of 60.0° to the door of Figure

8.7a, 2.00 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.

(b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door.

What minimum force must the wedge exert so that the force applied in part (a) won’t open the door?

4


Q5 : A uniform 35.0-kg beam of length L = 5.00 m is supported by a vertical rope located d = 1.20 m from its left end. The right end of the beam is supported by a vertical column. Find

(a) the tension in the rope and

(b) the force that the column exerts on the right end of the beam.

Q6 : Calculate the net torque (magnitude and direction) on the beam in the figure about:

(a) an axis through O perpendicular to the page and

(b) an axis through C perpendicular to the page.

5


Q7 : A vault is opened by applying a force of 300 N perpendicular to the plane of the door, 0.80 m from the hinges. Find the torque due to this force about an axis through the hinges. a. b. c. d.

120 N

 m

240 N

 m

300 N

 m

360 N

 m

Q8 : A 3.0-m rod is pivoted about its left end. A force of 6.0 N is applied perpendicular to the rod at a distance of 1.2 m from the pivot causing a counterclockwise ( ccw ) torque, and a force of 5.2 N is applied at the end of the rod 3.0 m from the pivot. The 5.2 N is at an angle of 30



to the rod and causes a cw torque. What is the net torque about the pivot? a. b. c. d.

15 N

 m

0 N

 m



6.3 N

 m



0.6 N

 m

Q9 : A uniform beam of length 4.0 m and weight 100 N is mounted on an axle at one end perpendicular to the length of the beam. A rope is attached to the end of the beam at the other end from the axle and the beam is lifted by the rope so that the beam makes an angle of 30° with the horizontal. What is the tension in the rope if it is straight up? a. b. c.

50 N

87 N

100 N d. d. 200 N

Q10 : A rod of length L is pivoted about its left end and has a force F applied perpendicular to the other end. The force F is now removed and another force F' is applied at the midpoint of the rod. If F' is at an angle of 30



with respect to the rod, what is its magnitude if the resulting torque is the same as when F was applied? a. b. c.

F

2F

3F

4F

6


8.2Torque and the Two Conditions for Equilibrium

An object in mechanical equilibrium must satisfy the following two conditions:

1. The net external force must be zero:

Where : F net

(x) = 0

F net

(y) = 0

F net

(z) = 0

The object is at rest or moving with a constant speed.

2. The net external torque must be zero:

Where clockwise torque = counterclockwise torque.

7


1. A woman of mass m = 55.0 kg sits on the left end of a seesaw a plank of length L =

4.00 m, pivoted in the middle as in the figure.

(a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 75.0 kg sit if the system (seesaw plus man and woman) is to be balanced?

(b) Find the normal force exerted by the pivot if the plank has a mass of m pl

=

12.0 kg.

(c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank.

8


2. find the location of a person’s center of gravity. Suppose your lab partner has a height L of 173 cm and a weight w of 715 N. You can determine the position of his center of gravity by having him stretch out on a uniform board supported at one end by a scale, as shown in the figure . If the board’s weight w b is 49 N and the scale reading F is 3.50 x 10 2 N, find the distance of your lab partner’s center of gravity from the left end of the board.

9


3. A 50.0-N bowling ball is held in a person’s hand with the forearm horizontal. The biceps muscle is attached 0.0300 m from the joint, and the ball is 0.350 m from the joint. Find the upward force F exerted by the biceps on the forearm (the ulna) and the downward force R exerted by the humerus on the forearm, acting at the joint. Neglect the weight of the forearm and slight deviation from the vertical of the biceps.

10


4. A uniform ladder 10.0 m long and weighing 50.0 N rests against a smooth vertical wall as in Figure a. If the ladder is just on the verge of slipp ing when it makes a 50.0° angle with the ground, find:

(a)The static friction force ( f ) between the ladder and the ground.

(b)The normal force ( n ) between the ladder and the ground.

5. A uniform horizontal beam 5.00 m long and weighing 3.00 x 10 2 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0° with the horizontal. If a person weighing 6.00 x 10 2

N stands 1.50 m from the wall, find : the magnitude of the tension T in the cable and the force R exerted by the wall

11


6.

A ladder 3 m long and weighing 80 N is leaning against a wall, as shown below. The ladder makes an angle of 30 0 to the horizontal. The vertical side of the wall is frictionless, but there is friction along the floor with m s

=

0.2.

maximum.

A 600 N person is standing on the ladder. a.

Calculate the value of the contact force where the ladder is touching the wall. b.

How far up the ladder can the person stand before the ladder will begin to slip.

12


Newton’s second Law of Rotational Motion: states that:

For a rigid body rotating about a fixed axis, the net torque

 acting on an object that has a moment of inertia I causes an angular acceleration α.

=

I α

The angular acceleration of an extended rigid object is proportional to the net torque acting on it.

---------------------------------------

1.

A toy consisting of two balls, each 0.45 kg, at the ends of a 0.46-m-long, thin, lightweight rod is shown in figure. Find: a.

the moment of inertia of the toy about the center of the rod resulted by one rigid object.

I = mr 2 = 0.45 x 0.23 = b.

The torque resulted if the linear acceleration is 1.5 m/s 2 for each object.

13


2.

The uniform rigid rod of mass m, length L, and rotational inertia I shown below is pivoted at its left-hand end. The rod is released from rest from a horizontal position.

What is the linear acceleration of the rods center of mass the moment after the rod is released?

A. mgL

2

√ B.

2 I mgL

2

C.

4 I mgL

2

I

D. mgL

E.

2 I

2 mgL

2

I

3.

A grinding wheel is a solid, uniform disk of mass 2.50 kg and radius 9.00 cm. Starting from rest, what constant torque must a motor supply so that the wheel attains a rotational speed of 126 rev/s in a time of 6.00 s?

4.

If you see an object rotating, is there necessarily net torque acting on it?

Explain.

No, not necessarily,

i.e, only if its angular velocity changes. The net torque equals to the moment of inertia multiplied by the angular acceleration, and thus for a constant angular velocity, there is no angular acceleration, and thus no net torque

14


Angular Momentum

A torque exerted on an object can change the angular momentum of an object. a. Angular momentum is a vector quantity, with its direction determined by a right-hand rule ( Align your right hand so that, as you curl your fingers in toward your palm, your fingertips follow the object’s rotation; then your thumb points in the direction of L .

b. The magnitude of angular momentum of a point object about an axis can be calculated by multiplying the perpendicular distance from the axis of rotation to the line of motion by the magnitude of linear momentum. c. The magnitude of angular momentum of an extended object can also be found by multiplying the rotational inertia ( I ) by the angular velocity ( ω )

L = I ω

Remember : I = mr 2 d. The change in angular momentum of an object is given by the product of the average torque and the time the torque is exerted.

∆t = ∆L = L f

– L i

= I ω f

- I ω i

= I ( ω f

– ω i

) = I ∆ ω

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1.

A turntable of mass 5.00 kg has a radius of 0.100 m and spins with a frequency of 0.550 rev/s. What is its angular momentum? Assume the turntable is a uniform disk.

2.

Assume the Earth is a uniform solid sphere with radius of 6.37 x 10 6 m and mass of 5.97 x 10 24 kg. Find the magnitude of the angular momentum of the Earth due to rotation about its axis.

3. The mass of a flywheel is 5.6 x10 4 kg. This particular flywheel has its mass concentrated at the rim of the wheel. If the radius of the wheel is 2.6 m and it is rotating at 350 rpm, what is the magnitude of its angular momentum?

16


4.

Six flywheels have masses, thicknesses, radii, and angular speeds as given in the table. Each flywheel is a solid disk. Rank the flywheels in order of their angular momentum, smallest to largest.

17


5.

The angular momentum of a spinning wheel is 240 kg ⋅ m 2 /s. After application of a constant braking torque for 2.5 s, it slows and has a new angular momentum of

115 kg ⋅ m 2 /s. What is the torque applied?

6.How long would a braking clockwise torque of 4.00 N ⋅ m have to act to just stop a spinning wheel that has an initial angular momentum of 6.40 kg ⋅ m 2 /s?

18


Conservation of Angular Momentum if the net external torque acting on a system is zero, then the angular momentum of the system cannot change. This is the law of conservation of angular momentum :

1. A student sits on a pivoted stool while holding a pair of weights as in the figure below. The stool is free to rotate about a vertical axis with negligible friction. The moment of inertia of student, weights, and stool is 3.25 kg. m2. The student is set in rotation with arms outstretched, making one complete turn every 1.5 s. As he rotates, he pulls the weights inward so that the new moment of inertia of the system (student, weights, and stool) becomes 2.80 kg. m2. What is the new angular speed of the system?

19


2. A 0.10-kg mouse is perched at point B on the rim of a 2.00-kg wagon wheel that rotates freely in a horizontal plane at 1.00 rev/s in the figure below. The mouse crawls to point A at the center. Assume the mass of the wheel is concentrated at the rim. What is the frequency of rotation in rev/s when the mouse arrives at point A ?

20


Angular Momentum in Planetary Orbits

Conservation of angular momentum applies to planets orbiting the Sun in elliptical orbits.

Kepler’s second law says that the orbital speed varies in such a way that the line from the Sun to the planet sweeps out area at a constant rate ( Fig. below ). You can show that Kepler’s second law is a direct result of conservation of angular momentum, where the angular momentum of the planet is calculated using an axis of rotation perpendicular to the plane of the orbit and passing through the Sun.

When the planet is closer to the Sun, it moves faster; when it is farther away, it moves more slowly.

Conservation of angular momentum can be used to relate the orbital speeds and radii at two different points in the orbit. The same applies to satellites and moons orbiting planets.

1. At perihelion (closest approach to the Sun), Earth is 1.47 . 108 km from the Sun and its orbital speed is 30.3 km/s. What is Earth’s orbital speed at aphelion

(greatest distance from the Sun), when it is 1.52 . 108 km from the Sun? Note that at these two points Earth’s velocity is perpendicular to a radial line from the

Sun.

21


2. An ice-skater, with a mass of 60.0 kg, glides in a circle of radius 1.4 m with a tangential speed of 6.0 m/s. A second skater, with a mass of 30.0 kg, glides on the same circular path with a tangential speed of 2.0 m/s. At an instant of time, both skaters grab the ends of a lightweight, rigid set of rods, set at 90 ° to each other, that can freely rotate about a pole, fixed in place on the ice.

(a) If each rod is 1.4 m long, what is the tangential speed of the skaters after they grab the rods?

(b) What is the direction of the angular momentum before and after the skaters

“collide” with the rods?

22


3. In a motor, a flywheel (solid disk of radius R and mass M ) is rotating with angular velocity w i . When the clutch is released, a second disk (radius r and mass m ) initially not rotating is brought into frictional contact with the flywheel.

The two disks spin around the same axle with frictionless bearings. After a short time, friction between the two disks brings them to a common angular velocity.

(a) Ignoring external influences, what is the final angular velocity?

(a) Strategy Use conservation of angular momentum, since no external torques act on the two-disk system.

Solution Find the final angular velocity.

L f



I f

 f



L i



I



, so

 f



I



I f



I i

I

 1

2 mr

2



1

 mr

2

 i

[2( MR

2



2)] 1



 i mr

2

( MR

2

)

.

(b) Does the total angular momentum of the two change? If so, account for the change. If not, explain why it does not.

(b)Strategy and Solution

The total angular momentum does not change, since no external torques act on the system.

23

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