The rotating systems shown in the figure differ only in that the two identical movable masses are positioned a distance r from the axis of rotation (left),or a distance r/2 from the axis of rotation (right).You release the hanging blocks simultaneously from rest,and call t_L the time taken by the block on the left and t_R the time taken by the block on the right to reach the bottom,respectively.The bar,pulley,and rope have negligible mass,the rope does not slip,and there is no friction in the axle of the pulley.Under these conditions A) t_L = t_R. B) t_L = t_R. C) t_L = t_R. D) t_L = 2t_R. E) t_l = 4t_R.

Question

Quizplus · Accepted Answer

The time it takes for a hanging mass to reach the bottom of a rotating system is given by the formula:

t = π√(r/g)

where r is the distance between the hanging mass and the axis of rotation, and g is the acceleration due to gravity.

For the left system, the distance between the hanging mass and the axis of rotation is r, so the time it takes to reach the bottom is:

t_left = π√(r/g)

For the right system, the distance between the hanging mass and the axis of rotation is r/2, so the time it takes to reach the bottom is:

t_right = π√(r/2g)

Dividing t_left by t_right gives:

t_left/t_right = √(r/(2r)) = √2

Therefore, t_left = √2 t_right = 2t_right.

So the answer is choice D.

The Rotating Systems Shown in the Figure Differ Only in That