# Determine the angular acceleration of the block

1. Nov 6, 2005

### ViewtifulBeau

A 6.59 kg block is released from A on a frictionless track, that slopes down until it reaches the floor, then loops in a circle(kinda like the hotwheels loops). Determine the angular acceleration of the block, alpha, about O, the axis of the loop, at the point P, which is located 45 degrees counterclockwise from the floor(the line connecting point P and O is parallel to the table). Take the positive direction into the page. The values of the variables given in the figure are: h = 8.4 m, R = 2.4 m. (Hint: You may not need all the information given.)

sorry about the discription, a picture is given.

I found the angular velocity of the block by finding the velocity of the block with sqrt(gh)= 10.849 with h = 8.4-6.4 then finding the circumference of the loop 2piR= 15.0786.

then w = 10.8/15.1 = .719448. i know that angular accel is the change dw/dt and i think that gravity will play a part in the acceleration, but I don't know how. thanks

2. Nov 7, 2005

### Fermat

The body is moving with non-uniform circular motion, so use the transverse component of acceleration, which is given by,

$$mr\ddot\theta = mgsin\theta$$

$$\mbox{where } \theta \mbox{ is the angle subtended by the radial line to the block, and the perpindicular from O to the floor.}$$

So,

$$\ddot\theta = \frac{g}{r}sin\theta$$
$$\mbox{when } \theta = 90^o,\ R = 2.4,$$
$$\ddot\theta = \frac{9.8}{2.4}sin 90$$
$$\ddot\theta = 4.08\ rad/sec^2$$

As the hint suggested, you may not need all of the info given

Last edited: Nov 7, 2005