# Homework Help: Rotational Problem

1. Dec 19, 2005

### Eumcoz

Hey, I have a problem...don't really know how to tackle it.

A Uniform 6.0m long rod of mass 50kg is supported b y a wire at one end.(Other end of wire is connected to wall) The other end rests against a wall where it is held by friction. The coefficient of static friction between the wall and the rod is .6. A 20kg mass is suspended some distance,x, from the wall. The wire makes a 37 degree angle with the rod.

(a) What is the minimum distance, x, that the mass must be from the wall(Point a) in order that the rod does not slip at the point at the wall.

(b) Calculate the total moment of inertia of the rod plus weight about the point where the rod and wire connect(Point B)(Assume the 20kg mass is a point particle at a distance x from the wall and a distance 6.0m - x from point b)

(c) If the wall was to get slick, such that the coefficient of static friction was reduced to 0, what would be the inital angular acceleration of the rod and weight around the point b?(You will need to find the sum of the torques around point b, with no torque being exerted on the wall.)

(d) What is the tangential linear aceleration of the 20kg mass?

There is a figure...it looks like a right triangle, with a mass hanging down if you can't picture the figure.

I figure(not sure) that the sum of the torque has to be equal to 0 for the first part. But do i have to factor in the Friction force at all? If not i got x = 5.13m, can anyone confirm this number?.

For the second part, i think i would use formula I = (Sum of) mr^2...so (50kg * 6m^2 * .5) + (20kg * x^2(.87m^2)) = 900 kgm^2 + 15.138 kgm^2 = 915.138 kgm^2 and this is around point b...so that would make the torque of the wire = 0 since r = 0 correct?

For the third part i would find the sum of the torque around point b...so ((6m*.5 * 9.8 m/s^2)(50kg)) + (20kg * x^2(.87m^2) * 9.8 m/s^2)=8820 N*m +170.52 N*m = 8990.52 Nm and then use that to figure out the Angular Accel using the formula Aa(angular accel.) = Torque / Moment Inertia so... 8990.52 N*m / 915.138 kgm^2 = 9.824 rad/s^2

I am a little iffy on the last part. Would the accel of the weight be the same as the whole system? So would you just do At(tangental accel) = .87m * 9.824rad/s^2 = 8.54 m/s^2? or is this step wrong?

Any help on this problem will be greatly appriciated...thanks in advance!

Last edited: Dec 19, 2005
2. Dec 19, 2005

### Eumcoz

3. Dec 21, 2005

### mukundpa

It depend on the axis of rotation you are taking.

If you are taking torque about the end with wire, you have to take torque due to friction, but if you are taking torque about the end touches the wall torque due to friction is zero, but torque due to tension in the wire will come in picture.

For equilibrium

1. Sum of all the forces on the body must be zero and
2. Sum of the torque must be zero.