Part A: A car with 56-cm -diameter tires accelerates uniformly from rest to 20 m

Question

Part A: A car with 56-cm -diameter tires accelerates uniformly from rest to 20 m/s in 20 s . How many times does each tire rotate?

Part B: An athlete at the gym holds a 2.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? What is the magnitude of the torque about his shoulder if he holds his arm straight, but 60 below horizontal?

Part C:

If you are using a wrench to loosen a very stubborn nut, you can make the job easier by using a "cheater pipe." This is a piece of pipe that slides over the handle of the wrench, making it effectively much longer. Explain why this would help you loosen the nut.

The cheater pipe ___ the wrench in the radial direction, providing a _____ moment arm for the force you exert.

For a given force the torque exerted with the cheater pipe is _______

extends

shortens

larger

smaller

Explanation / Answer

angular accleration (A) = rate of change of angular velocity

A = (wf-Wi)/t

Velocity V = r W

W f = V/r = 20/0.28

Wf = 71.42 rad/s

so

A = ( 71.42 -0)/(20)

A = 3.57 rad/s^2

2 A Theta = wf^2

theta = 2pi n

n = (71.4^2)/(2 * 3.57*2* 3.14)

n = 114 turns

------------------------------
As sum of torque about his shoulder is given by

T = g*0.7*(2+4/2)

T = 27.44 N m

now at angle of 60 degrees

sum of torque= 9.8*0.7*(2+4/2)*cos60

T = 13.72 N m

=------------------------

Torque = R x F or R F sin theta

because R= Length of wrench + rod generating the greatest torque

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