Centripetal acceleration and KE problem

[completelyme]

The question and information given is -
"The formula for the centripetal acceleration experienced by a body moving at speed v in a circle of radius r is ac = v squared / r

A small mass m slides without friction along a loop-the-loop track. The circular loop has radius R. The mass starts from rest at point P a distance h abouve the bottom of the loop.

a. What is the KE of m when it reaches the top of the loop?
b. What is its acceleration at the top of the loop assuming that it stays on the track?
c. What is the least value of h if m is to reach the top of the loop without leaving the track?"

PE = mgh at point P, so that would be 9.8mh, and KE = 0
a. At the top of the loop, how would you figure out the KE? If KE = 1/2mv2, it would be 1/2m and v2 = acr, how would you express this?
the PE = mg2R, so how would you use this to find the KE?
b. How would the ac equation be used to figure this out?
c. Would this answer be 2R, since objects return to their original height in the absence of friction? Or does the loop complicate this?

I'm sorry about this question, but finals are this week and my brain's slightly fried.

Thanks,
Ashlee

 

[lightgrav]

Obviously, the KE remaining when at the top of the track depends on "h".

did you draw a free-body diagram of the small mass at the top of the loop?
What direction are the Forces? These cause the acceleration.
(you now know a formula for the acceleration component that is perp. to v).

To JUST stay in contact with the track, F_Normal = 0 .

 

[quicknote]

I completely understand the pressure and stress of finals week. Let me break down the problem and provide some guidance to help you answer the questions.

First, let's start with the formula for centripetal acceleration, which is ac = v^2 / r. This formula tells us that the acceleration of an object moving in a circle is directly proportional to the square of its velocity and inversely proportional to the radius of the circle. This means that as the velocity increases, the acceleration increases, and as the radius decreases, the acceleration increases.

a. To find the KE of the mass at the top of the loop, we need to use the formula KE = 1/2mv^2. We know that the mass starts from rest at point P, so its initial velocity is 0. Therefore, at the top of the loop, its velocity will be its maximum, which can be calculated using the formula v^2 = acr. So, the KE at the top of the loop would be 1/2m(acr)^2.

b. At the top of the loop, the acceleration of the mass can be calculated using the formula ac = v^2 / r. We already know the velocity (calculated in part a) and the radius of the loop, so we can easily plug in the values and calculate the acceleration.

c. To find the minimum value of h, we need to ensure that the mass reaches the top of the loop without leaving the track. This means that the centripetal acceleration must be greater than or equal to the acceleration due to gravity (g) at the top of the loop. So, we can set up the inequality ac ≥ g and substitute the values for ac and g to solve for the minimum value of h.

I hope this helps you understand the problem better and gives you a starting point to solve it. Remember to always use the appropriate formulas and units to solve any physics problem accurately. Good luck on your finals!