# Rotation & Energy problem

1. Nov 3, 2013

### vac

1. The problem statement, all variables and given/known data
Torque = 3
Time = 3.12 s
Length of the each rod = 1 m so the radius = 0.500 m
mass of each rod is = 0.500 kg
$$M_1 = 4 kg$$
$$M_2 = 2 kg$$
$$M_3 = 4 kg$$
$$M_4 = 2 kg$$
http://imageshack.us/a/img27/5475/qu51.jpg [Broken]
2. Relevant equations
What is the energy of the object (please refer to the picture above) that has a mass fixed to each of its four corners? If it rotates for 3.12 s? Torque and time is given.
3. The attempt at a solution
$$K_E = \frac{1}{2} I \omega^2$$
$$I = \sum mr^2$$
$$I = (4+2+4+2+0.5+0.5)(0.5)^2$$
$$I = 3.25$$
$$\omega = \alpha t$$
Now we have "I" and Torque we can calculate alpha.
$$Torque = I \alpha$$
$$3 = 2.25 \alpha$$
$$\frac{3}{3.25}= \alpha$$
so
$$\omega = \alpha t$$
$$\omega = \frac{3}{3.25} 3.12 s = 2.88$$
substitute back in
$$K_E = \frac{1}{2} I \omega^2$$
$$K_E = \frac{1}{2} 3.25 (2.88)^2$$
$$K_E = 13.5 J$$
Can you please tell me what is wrong with my answer and if there is an easier way to solve such problems.
Thanks.

Last edited by a moderator: May 6, 2017
2. Nov 3, 2013

### haruspex

Pls explain how you get that. What is the value of Ʃm?

3. Nov 3, 2013

### vac

summation of masses times radius square: are (m1+m2+m3+m4+ mass of two rods )(r^2) = (4kg + 2kg + 4kg + 2kg + 0.5kg + 0.5 kg)(0.5 m)^2 = 3.25

4. Nov 3, 2013

### vac

Ok, do you see how I did that?

5. Nov 3, 2013

6. Nov 4, 2013

### vac

Thank you for asking this question, it is the main thing that drove me to make this post.
I think it should be (1/12) times mass times length of both rods ... (1/12)ML.
But how about the vertical rods?

7. Nov 4, 2013

### vac

The moment of inertia for a rod is $\frac{1}{12} ML^2$

8. Nov 4, 2013

### haruspex

Yes. You already included the vertical rods correctly. The 1/12 formula is for a rod rotating about its centre. Every part of each vertical rod is distance 0.5m from the axis, so 0.5m is right for those.