Exercise Solution

Part a. Concepts

 

1.  When is the  linear momentum of a system conserved?

 

When the net force acting on the system is 0

 

2.  When is the kinetic energy of a system conserved?

 

For collisions that are elastic

 

3.  When is total energy of a system conserved?

 

Always

 

4.  When is mechanical energy of a system conserved?

 

When only conservative forces are acting

 

5. When is total angular momentum of a system conserved?

 

When the net torque acting on the system is 0

 

6. Rotating rods - equal weight - demonstration. Clearly explain why it was more difficult to rotate one rod than the other.

 

Since I is given by mr², the rod with the mass near the center has a lower I if rotated about the center.

 

7. Rotating mass on a string demonstration. Why did the angular velocity increase as I pulled on the string?

 

Net torque acting on the 2-D system is 0, therefore angular momentum is conserved.

(Iw)_initial = (Iw)_final            (mr²w)_initial = (mr²w)_final

The mass did not change but r decreased.

In order to compensate - for the conservation of angular momentum, w had to increase

 

8. Why do animals that can run very fast have most of the mass of their legs closer to their body?

 

When the mass of the leg is concentrated next to the body, the leg has a lower moment of inertia (I = mr²).

The result is that the legs will require less torque for a given angular acceleration. This means that they can develop a higher angular acceleration and run faster.

 

9. Is it possible for a non-rigid body to have a single value of the angular velocity?

 

No. Since the body is non-rigid, the angular position of one part of the body changes with respect to other parts of the body. 

 

10. Can a small force ever exert a greater torque than a larger force?

Yes. Since t = F_perpr      a smaller F can result in a larger torque if the radius is sufficiently large

The angle also plays a part.

 

Part b: Problems

 

Problem 4

 

First, we find the initial angular velocity

w₀ = (6500 rev/min)x(2p rad/rev)x((1 min/60 sec)  = 681 rad/s

 

a = Dw/Dt = (0 – 681 rad/sec)/3.0 s  227 rad/s²

 

Problem 15

The angular displacement can be found from the following uniform angular acceleration relationship.

 

q = (1/2)x(w₀ - w)t = (1/2)x(0 + 1500 rev/min)x(220 s)x((1 min/60 s) = 2.8 x 10⁴ rev  or 28000 rev

                       

Problem 22a

The maximum torque will be exerted by the force of her weight, pushing tangential to the circle

in which the pedal moves.

 

t =  r_perpF = r_permg = (0.17 m)x(55 kg)x(9.8 m/s²) = 92 mN

                                   

Problem 22b

She could exert more torque by pushing down harder with her legs, raising her center of mass. She could also pull upwards on the handle bars as she pedals, which will increase the downward force of her legs

 

Problem 27

Using the table, I = (2/5)MR² = 1.81 kg m²

 

Problem 43

The energy from rest to its final speed is just the final rotational KE of the rotor.

 

KE_rot = (1/2)Iw² = (1/2)x(3.75x10-² kg m²)x ((8250 rev/min)x(2prad/1 rev)x(1 min/60sec))2 = 1.40x10⁴ or 14000 J

           

Problem 51

L = Iw = = MR²w = (0.210 kg)x(1.10 m)²x(10.4 rad/s) = 2.64 kg m²/s