Rotational Dynamics

M=mass of the wheel= 11 Kg
R=radius of the wheel= 0.7 m
a=acceleration of the centre of mass of the wheel= 0.55 m/s^2

(a) What are the magnitude and direction of the frictional force on the wheel?
Magnitude: ______N

Let the magnitude of the frictional force= f
The direction of the frictional force would be to the left (opposite to the 10 N force)

Net force acting to the right= 10 - f

Since force = Mass x Accelerarion
10 - f= M a
But M a= 6.05 N =11x0.55

Therefore f= 10 -M a= 3.95 N =10-6.05

Answer: Magnitude= 3.95 N
Direction: left

(b) What is the rotational inertia of the wheel about an axis through its center of mass and perpendicular to the plane of the wheel? ______kg ? m^2

alpha =angular acceleration=a/R= 0.7857 rad/s^2 =0.55/0.7

Torque=T=I alpha
where
I = rotational inertia
alpha = angular acceleration

Torque is provided by the frictional force
T= f R
f= 3.95 N
R= 0.7 m
Therefore T= 2.765 N-m =3.95*0.7

Since T= I alpha
and alpha = 0.7857 rad/s^2
I= 3.5192 kg ? m^2 =2.765/0.7857

Answer:
rotational inertia of the wheel about an axis through its center of mass and perpendicular to the plane of the wheel=
3.5192 kg ? m^2

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