A small disk can slide in a circular path on | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $l$ be the length of the thread. When the disk is released from the horizontal position,it moves to the lowest point of its circular path on t

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) Let $l$ be the length of the thread. When the disk is released from the horizontal position,it moves to the lowest point of its circular path on the inclined plane.In the lowest position,the disk has descended a vertical height $h = l \sin 30^\circ = l/2$.Applying the work-energy theorem,the change in kinetic energy equals the work done by gravity:$\frac{1}{2}mv^2 = mg(l \sin 30^\circ) = mg(l/2) = \frac{mgl}{2}$.Thus,$mv^2 = mgl$.At the lowest point,the forces acting along the radial direction are the tension $T$ (towards the center) and the component of gravity $mg \sin 30^\circ$ (away from the center).The net centripetal force is $T - mg \sin 30^\circ = \frac{mv^2}{l}$.Substituting $mv^2 = mgl$ and $\sin 30^\circ = 0.5$:$T - mg(0.5) = \frac{mgl}{l} = mg$.$T = mg + 0.5mg = 1.5mg$.Therefore,the tension is $1.5mg$.

Similar Questions

The friction coefficient between the horizontal surface and each of the blocks shown in the figure is $\mu = 0.2$. The collision between the blocks is perfectly elastic. What is the separation between the blocks when they come to rest? (in $cm$)

$A$ particle of mass $m$ starts moving from the origin along the $x$-axis and its velocity varies with position $x$ as $v = k \sqrt{x}$. The work done by the force acting on it during the first $t$ seconds is ...........

$A$ body of mass $2 \,kg$ thrown vertically from the ground with a velocity of $8 \,ms^{-1}$ reaches a maximum height of $3 \,m$. The work done by the air resistance is (acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,J$)

$A$ free body of mass $8 \, kg$ is travelling at $2 \, m/s$ in a straight line. At a certain instant,the body splits into two equal parts due to an internal explosion which releases $16 \, J$ of energy. Neither part leaves the original line of motion. Finally:

$A$ body of mass $5\,kg$ is moving with a momentum of $10\,kg\,m/s$. Now a force of $2\,N$ acts on the body in the direction of its motion for $5\,s$. The increase in the kinetic energy of the body is $...........J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module