A torque of 0.97 N m is applied to a bicycle wheel of radius 35 cm and mass 0.75 kg. Assuming the wheel can be modelled as a hoop, find the angular acceleration of the wheel.

Explanation:

Given

torque,τ= 0.97N

radius r= 35cm to m= 0.35m

mass m= 0.75kg

Required

angular acceleration

First, we need the  moment of inertial of the wheel

I=mr^2

I=0.75*0.35^2

I=0.75*0.1225

I=0.09kgm^2

Secondly, the angular moment of inertial is expressed as

a=τ/I

a=0.97/0.09

a=10.77rads/s^2

a=11 rad/s^2 approx.

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000