Angular acceleration homework problem

AI Thread Summary
The discussion revolves around calculating the tangential acceleration, tangential velocity, and centripetal acceleration of a bug on a rotating disk. The bug is on an 8.0-inch diameter disk that accelerates to 75 revolutions per minute in 5 seconds. The tangential acceleration was calculated as 0.1596 m/s² and the tangential velocity as 0.79796 m/s. The centripetal acceleration is derived using the formula ac = v²/r, where the radius is converted to meters. Participants confirm the calculations and seek clarification on the correct approach to find the centripetal acceleration.
Dr_bug
Messages
16
Reaction score
0

Homework Statement


What is the tangential acceleration of a bug on the rim of a 8.0 in. diameter disk if the disk moves from rest to an angular speed of 75 revolutions per minute in 5.0 s?
(b) When the disk is at its final speed, what is the tangential velocity of the bug?

(c) One second after the bug starts from rest, what is its tangential acceleration?

What is its centripetal acceleration?

What is its total acceleration (magnitude and angle relative to the tangential acceleration) ?

Homework Equations


atan= R*angular acceleration
v= R*w
ac=R*w^2

The Attempt at a Solution


I was able to answer a, b, and the first part of c but I can't get the centripetal acceleration . For (a) I got .1596 m/sec^2 (b) .79796 m/ sec (c) the atan after 1 sec is the same as (a). To find the ac I used the third equation but didn't get the right answer... is that the right equation to use to find centripetal acceleration? I converted 4 in into .1016 m and used that for R and then multiplied it by 7.85 rad/sec^s... i could use some help
 
Physics news on Phys.org
I don't agree with your (a) and (b) answers . . . can you show the calcs?
Once you have the v value from (b) it should be easy to find a = v²/r.
 
sure (a):
(4 in* 2.54 cm*75 rev*2pi rad)/(100 cm*60 sec*5 sec)= .1596 m/sec^2 -- I know its right because I already submitted it online and it accepted the answer.
(b):
(4 in* 2.54 cm*75 rev*2pi rad)/(100 cm*60 sec)= .79796 m/s -- this was also accepted
 
Oh, I didn't notice the inches! Agree.
 
so should I then take .79796 and square it and then divide by .1016 (thats the radius converted to meters)
 
Right!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top