Question

A smooth ring A of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block C of mass M(= 2m) as shown in the Figure. At an instant the string between the ring and pulley makes an angle θ with the rod.

(a) Show that, if the ring slides with a speed v, the block descends with speed v cos θ .

(b) With what acceleration will the ring start moving if the system is released from rest with θ = 30°?

Answer 1

Find the constraint relation between the acceleration of the ring and the block. Measure the distances of ring and the block from the fixed pulley B.

(a) Suppose in a small time interval ∆t the ring is displaced from A to A′ and the block from C to C′. Drop a perpendicular A′P from A′ to AB. For small displacement A′B = PB. Since the length of the string is constant,

we have AB + BC = A'B + BC'

or, AP + PB + BC = A'B + BC

or, AP = B'C - BC = CC' (as A'B = PB)

or, AA'cosθ = CC'

or, \(\frac{AA'\cos\theta}{\Delta t} = \frac{CC'}{\Delta t}\)

or, (velocity of the ring) cosθ = (velocity of the block).

(b) If the initial acceleration of the ring is a, that of the block will be a cosθ . Let T be the tension in the string at this instant. Consider the block as the system. The forces acting on the block are

(i) Mg downward due to earth, and

(ii) T upward due to string.

equation of motion of the block is

Mg-T = Macosθ   ... (i)

Now, consider the ring as the system. The forces acting on the ring are

(i) Mg downward due to gravity,

(ii) N upward due to the rod,

(iii) T along the string due to string.

Taking components along the rod, the equation of motion of the ring is

Tcosθ = ma   ......(ii) 

From (i) and (ii),