A wheel of radius R rolls without slipping with angular speed $\omega$,the linear speed of its center is A.The masses and the radii of discs A and B are respectively 2 kg and 0.2 kg,and 25 cm and 2.5 cm.What is the moment of inertia of the composite object? A)0. B)$\omega$/R. C)$\omega$R. D)$\omega$R².

Question

Quizplus · Accepted Answer

The linear speed of the center of the wheel is given by: 
v = ωR 
where ω is the angular speed and R is the radius of the wheel. 
Using the given values: 
v = (F1F1F1S1F1F1F10) * R = F1F1F1S1F1F1F10R 
The moment of inertia of the composite object can be found by adding the moment of inertia of the wheel and the two discs: 
I = Iwheel + Ia + Ib 
The moment of inertia of the wheel is given by: 
Iwheel = (1/2) MR^2 
where M is the mass of the wheel. 
Using the given values: 
Iwheel = (1/2) * 2 * R^2 = 2R^2 
The moment of inertia of each disc is given by: 
I = (1/2) mr^2 
where m is the mass of the disc and r is its radius. 
Using the given values: 
Ia = (1/2) * 0.2 * (0.25)^2 = 0.00625 
Ib = (1/2) * 0.2 * (0.025)^2 = 0.00003125 
Adding these up: 
I = 2R^2 + 0.00625 + 0.00003125 = 2R^2 + 0.00628125 
Substituting for v and simplifying: 
I = (mv^2) / (2ω^2) = (m/F1F1F1S1F1F1F10R^2) * v^2 = (m/F1F1F1S1F1F1F10R) * (F1F1F1S1F1F1F10R)^2 = F1F1F1S1F1F1F10R 
Therefore, the answer is C.

A Wheel of Radius R Rolls Without Slipping with Angular ω\omegaω

A Wheel of Radius R Rolls Without Slipping with Angular $\omega$