Subjective Type
In figure, a constant horizontal force $$\vec F_{app}$$ of magnitude of $$10\ N$$ is applied to a wheel of mass $$10\ kg$$ and radius $$0.30\ m$$. The wheel rools smoothly on the horizontal surface, and the acceleration of its centre of mass has magnitude $$0.60\ m/s^2$$.
What is the rotational inertia of the wheel about the rotation axis through its centre of mass?
Solution
"The direction of the frictional force will be in the backward direction of the motion of the wheel.
10−f=ma10-f=ma
Substituting m=10kgm=10kg and a=0.60m/s2a=0.60m/s2 in the above equation, we get
$$F_r=4N$$Fr=4N
Equation of motion for rotation of wheel
fr×R=Iαfr×R=Iα
Which gives
$$I = f_r * \dfrac{R *4*0.3}{\alpha*2}$$
$$I = 0.6 Kgm^2$$"
Rotational Dynamics
A bowler throws a bowling a lane. The ball slides on the lane with initial speed $$v_{com.0}=8.5\ m/s$$ and initial angular speed $$\omega _0 =0$$. The coefficient of kinetic friction between the ball and the lane is $$0.21$$. The kinetic friction force $$\vec f_{k}$$ acting on the ball causes an angular acceleration of the ball. When speed $$v_{com}$$ has decreases enough and angular speed $$\omega$$ has increased enough, the ball stops sliding and then rolls smoothly.
Angular acceleration?
Rotational Dynamics
The center of mass of the disc undergoes simple harmonic motion with angular frequency $$ \omega $$ equal to
