A horizontal plane supports a stationary vertical cylinder of rad… - Vidyadip

MCQ

A horizontal plane supports a stationary vertical cylinder of radius $R = 1\ m$ and a disc $A$ attached to the cylinder by a horizontal thread $AB$ of length $l_0 = 2\ m$ (seen in figure, top view). An intial velocity ($v_0 = 1\ m/s$) is imparted $AB$ to the disc as shown in figure. .......... $\sec$ long will it move along the plane until it strikes against the cylinder ? (All surface are assumed to be smooth)

A
$1$
✓
$2$
C
$5$
D
$3$

✓

Answer

Correct option: B.

$2$

b
since the tension is always perpendicular to the velocity vector, the work done by the tension force will be zero. Hence, according to the work energy theorem, the kinetic energy or velocity of the disc will remain constant during it's motion. Hence, the sought time $t=\frac{s}{v_{0}},$ where $s$ is the total distance traversed by the small disc during it's motion.

Now, at an arbitary position $(Fig.)$ $d s=\left(l_{0}-R \theta\right) d \theta$

so, $s=\int_{0}^{{l/R}}\left(l_{0}-R \theta\right) d \theta$

$\mathbf{o r}$

$s=\frac{l_{0}^{2}}{R}-\frac{R l_{0}^{2}}{2 R^{2}}=\frac{l_{0}^{2}}{2 R}$

Hence, the required time, $t=\frac{l_{0}^{2}}{2 R v_{0}}$

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