You turn off your electric drill and find that the time interval for the rotating bit to come to rest due to frictional torque in the drill is t. You replace the bit with a larger one that results in a doubling of the moment of inertia of the drill's entire rotating mechanism. When this larger bit is rotated at the same angular speed as the first and the drill is turned off, the frictional torque remains the same as that for the previous situation. What is the time interval for this second bit to come to rest?

where I is the moment of inertia of the drill, Δw is the change in the angular speed of the drill bit and Δt is the time required for the rotating bit to come to rest.

Now when the bit is replaced with a larger one that doubles the moment of inertia of the drill's entire rotating mechanism the frictional torque is

τ = I'(Δw/Δt') .................(equation 2)

where I' is the moment of inertia of the longer drillbit and Δt' is the time required for the rotating longer bit to come to rest.

Since the frictional torque remains the same as that for the previous situation, we equate equation 1 and equation 2 :

I(Δw/Δt) = I'(Δw/Δt')

Also remember that the replaced bit doubled the moment of inertia of the drill's entire rotating mechanism I' =2I:

Therefore I(Δw/Δt) = 2I(Δw/Δt')

When we calculate for Δt'

Δt' = 2Δt

Therefore, the time interval for this second bit to come to rest is 2Δt.

topeadeniran2 topeadeniran2 · Answer 2 · 2021-11-18T18:23:33+01:00

The time interval for this second bit to come to rest will be 2Δt.

Based on the information given, one had to calculate the friction torque that's acting on the drill bit. This will go thus:

τ = I(Δw/Δt) ................ equation 1

where,

I = the moment of inertia of the drill.
Δw = the change in the angular speed of the drill bit.
Δt = the time required for the rotating bit to come to rest.

Now, in a situation whereby the bit is replaced with a larger one, the moment of inertia will be calculated thus:

τ = I'(Δw/Δt') .............. equation 2

Since the frictional torque remains the same as that for the previous situation, then it can be inferred that we will equate equation 1 and equation 2 which will be:

I(Δw/Δt) = I'(Δw/Δt')

Therefore I(Δw/Δt) = 2I(Δw/Δt')

We will then calculate for Δt' which will be:

Δt' = 2Δt

Therefore, the correct option will be 2Δt.

Read related link on:

https://brainly.com/question/14479044