Question

A uniform rod has a length of 20cm and a mass of 1.5kg. I plan to spin the rod on my finger about its center. The rod will start from rest, and I want it to reach an angular speed of 4 revolutions per second. I plan to get it there by exerting a constant torque over a period of 1/4 of a rotation.

1. What torque should I apply?

2. Suggest two possible forces that I could apply to generate this torque.

3. How much work would I have to do on the rod to make this happen?

Answer 1

L = 20 cm , m = 1.5 kg,

Moment of inertia of rod about its center

I = (1/12)mL^2

I =(1/12)1.5*0.2^2 =0.005 kg.m^2

wo =0 ,w = 4 rev/s = 4*2*3.14 rad/s

w = 25.12 rad/s

theta = (1/4)rev = 2*3.14/4 =1.57 rad

from rotaitonal kinematic equation

w^2 -wo^2 = 2*alpha*theta

25.12^2 -0 = 2*alpha*1.57

alpha = 200.96 rad/s^2

1) Torque = I*alpha

= 0.005*200.96 =1.0048 N.m

2) The possible two forces are applied on two end of rod

3) from work energy thorem

W = (1/2)Iw2^2 - (1/2)Iw1^2

W = 0.5*0.005*25.12^2

W = 1.578 J