of 8 As part of a camival game, a 0.603-kg ball is thrown at a stack of 18.3-am

Question

of 8 As part of a camival game, a 0.603-kg ball is thrown at a stack of 18.3-am tal, 0.363-kg objects and hits with a perfectly velocity of 10.8 m/s Suppose the ball strikes the topmost object as shown to the right. Immediately after the collision, the ball has a horizontal velocity of 4.35 m/s in the same direction, the topmost object now has an angular velocity of 3.23 rads about its center of mass and all the objects below are undisturbed. If the objects center of mass is located 12.8 cm below the point where the ball hits, what is the moment of inertia ofthe object about its mass? 9.65 kg m What is the center of mass velocity of the tall object immediately after it is struck? Number m/s 2.06 Incorrect. Previous Give Up 8 view Solution a Try Again 0 Next Exit

Explanation / Answer

A:- To calculate Moment of inertia,I:-

We know that I= M*R^2 where M= Mass & R= radius along which it rotates.

R= 12.8 cm= 0.128 m
So, I= 0.363*0.128 = 5.94 * 10^(-3) kg-m^2= 0.0059 kg-m^2.

B:- To calculate velocity of center of mass of object:-

Anguler velocity is given as, Omega= 3.23 rad/s
So, 3.23= v/r where, v= velocity of center of mass of object & r= radius along which it rotates.
So, v= 3.23*0.128= 0.413 m/s.
So this is the required velocity of center of mass of object.

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