Explore BrainMass

# Rotational Dynamics

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

** 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

https://brainmass.com/physics/torques/physics-rotational-dynamics-17564

## 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

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