# Rotational Motion and Equilibrium Problems, Work Checking

1. Oct 25, 2005

### kris24tf

I have a few Rotational Motion and Equilibrium problems that I hav been working on but I am not sure if I completed them correctly. I will show what I have done so far and if anyone could point me in the right direction I would appreciate it. Please, numerical explanations are easier for me to understand.

1) A variation of Russell traction supports the lower leg in a cast. Suppose the leg and cast have a total mass of 15 kg and m1 is 4.5 kg. a) what is the reaction force of the leg muscles to the traction?

For this I took m1=F/2g, 4.5kg=F/19.6, so F=88.2N?

b) what must m2 be to keep the leg horizontal?

For this I took m1g+m2g-15kg(g)=0, m2=15-4.5kg=10.5kg

2)Two masses are supended by a pulley (Atwood Machine). The pulley itself has a mass of .2kg, radius of .15m and constant torque of .35mN. What is the magnitude of the acceleration of the suspended masses if m1=.4kg and m2=.8kg?

I don't kno how to show a picture here, so I'll try to describe it. It is an atwood machine with m2 on the left with forces m2g downward and T2 upward, m1 with m1g downward and T1 upward, and a pulley with R, tf, and T2 and T1 downward on m2 and m1 respectively.

I ended up with an equation of a/2=(m2-m1)g/m1+m2+M/2+T and I got 1.2 m/s^s which is the correct answer but I have a feeling my equation might be incorrect. Any advice on what equations to go off of here would help

3)An ice skater spinning with outstretched arms has an angular speed of 4 rad/s. She tucks her arms and decreases moment of inertia by 7.5%. a) what is the resulting angular speed?

b) By what factor does the skater's kinetic energy change?

This is where I am confused most. I know I should use the rotational Work Energy theorem somehow, but I don't know how...

Any advice on these woudl be great. Thank you for looking!

Last edited: Oct 25, 2005
2. Oct 26, 2005

### kris24tf

Anyone?

Hello?

3. Oct 26, 2005

### andrevdh

kris24tf,
concerning problem 2 - some guidelines:
The friction in the pulley increases the tension on the downwards moving side
and decreases it on the upwards moving side with the magnitude of the friction. That is if the tension would be T without any friction it will be
$$T_2=T+T_f$$
and
$$T_1=T-T_f$$
The resulting torque of these two tensions on the pulley will cause an angular acceleration according to
$$\tau=I\alpha$$
The connection between the acceleration of the masses and that of the pulley
is that it's tangential acceleration is given by
$$a=a_t=r\alpha$$

Last edited: Oct 26, 2005