Question

You hold a small ice cube near the top edge of a hemispherical bowl of radius 10.0 cm. You release the cube from rest. What is the magnitude of its acceleration at the instant it reaches the bottom of the bowl? Ignore friction. [Hint: you may want to find the speed at the bottom first---try using energy.]

Answer 1

To solve this problem it is necessary to apply the concepts related to energy conservation as well as centripetal acceleration.

By conserving energy we know that

`\Delta KE = \Delta PE`

Where,

KE = Kinetic Energy

PE = Potential Energy

`\Delta KE = \Delta PE`

`(1)/(2)mv_f^2-(1)/(2)mv_i^2 = mgh_i-mgh_f`

Initial Kinetic Energy according the statement is zero, same as final potential energí, therefore

`(1)/(2)mv_f^2 = mgh_i`

Re-arrange for v,

`v_f = √(2gh_i)`

Where h here represent the radius of hemispherical bowl.

We have also the definition of centripetal acceleration, which is

`a_c = (v^2)/(R)`

But we have that the radius is equal to the height, then

`a_c = (v^2)/(h_i)`

Replacing the previous value of velocity found,

`a_c = ((√(2gh_i))^2)/(h_i)`

`a_c = (√(2gh_i))/(h_i)`

`a_c = 2g`

Substituting the value for gravitational acceleration

`a_c = 2*9.8`

`a_c = 19.6m/s^2`

Therefore the radial acceleration of ice cube at bottom is
`19.6m/s^2`