A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bottom of the hill at 2.5 m. What is the snowball's change in centripetal acceleration from the top of the hill and the bottom of the hill?

Question

Oscar Randall

Physics

23 April, 11:33

A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bottom of the hill at 2.5 m. What is the snowball's change in centripetal acceleration from the top of the hill and the bottom of the hill?

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Conway · Answer 1

To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:

Centripetal Acceleration = v squared divided by r

where:

v = velocity, m/s

r = radium, m

assuming the velocity does not change:

at the top of the hill:

centripetal acceleration = (4.5 m/s^2) divided by 0.25 m

= 81 m/s^2

at the bottom of the hill:

centripetal acceleration = (4.5 m/s^2) divided by 1.25 m

= 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.

change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2

= 64.8 m/s^2 or 65 m/s^2