# Rotational Dynamics

** Please see the attached file for the related illustration **

A constant horizontal force of 10 N is applied to a wheel of mass 11 kg and radius 0.70 m as shown in Fig. 12-31. The wheel rolls without slipping on the horizontal surface, and the acceleration of its center of mass is 0.55 m/s2.

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

Magnitude: ______N

Direction :to the right or upwards or downwards or to the 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

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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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