Determine the angular acceleration of the block

[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

 

[Fermat]

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

[tex]mr\ddot\theta = mgsin\theta[/tex]

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

So,

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

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

 

[quicknote]

I would approach this problem by using the principles of rotational motion and Newton's laws of motion.

First, I would calculate the potential energy of the block at point A, which is given by mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of point A above the ground. In this case, h = 8.4 m and m = 6.59 kg, so the potential energy is 6.59 * 9.8 * 8.4 = 551.32 J.

Next, I would use the conservation of energy principle to determine the kinetic energy of the block at point P, which is equal to the potential energy at point A. This can be calculated using the formula KE = 1/2 * I * w^2, where I is the moment of inertia of the block and w is the angular velocity.

To find the moment of inertia, I would use the formula I = mr^2, where m is the mass of the block and r is the radius of the loop. In this case, m = 6.59 kg and r = 2.4 m, so I = 6.59 * 2.4^2 = 37.872 kg*m^2.

Substituting this into the kinetic energy formula, we get 551.32 = 1/2 * 37.872 * w^2, which can be rearranged to solve for w. This gives us w = sqrt(551.32 / 18.936) = 4.513 rad/s.

Finally, to find the angular acceleration, we can use the formula alpha = dw/dt. Since we already know w, we just need to find the time it takes for the block to travel from point A to point P. This can be calculated using the formula t = sqrt(2h/g), where h is the height difference between the two points and g is the acceleration due to gravity. In this case, h = 8.4 - 6.4 = 2 m, so t = sqrt(2 * 2 / 9.8) = 0.632 s.

Substituting this into the formula for alpha, we get alpha = 4.513 / 0.632 = 7.140 rad/s^2.

So, the angular acceleration of the block as it