Chapter 6: Q54P (page 175)

A puck of mass $m_{1}$ is tied to a string and allowed to revolve in a circle of radius R on a frictionless, horizontal table. The other end of the string passes through a small hole in the centre of the table, and an object of massis tied to it (Fig. above). The suspended object remains in equilibrium while the puck on the tabletop revolves. Find symbolic expressions for (a) the tension in the string, (b) the radial force acting on the puck, and (c) the speed of the puck. (d) Qualitatively describe what will happen in the motion of the puck if the value of $m_{2}$ is increased by placing a small additional load on the puck. (e) Qualitatively describe what will happen in the motion of the puck if the value of $m_{2}$ is instead decreased by removing a part of the hanging load.

Short Answer

Expert verified

(a) The expression for the tension in the string is T = $m_{2}$ g.

(b) The radial force acting on the puck is $F_{R} = m_{2} g$ .

(d) The mass $m_{2}$ will move in downward direction and the mass $m_{1}$ will move in smaller circular radius.

(e) The mass $m_{2}$ will move in upward direction and the mass $m_{1}$ will move in larger circular radius.

Step by step solution

Gravitational force

Gravitational force is the force exerted by the earth on object. This force is directed towards the centre of the earth.

The expression for the gravitational force is given by

$F = m g$

Here F is the gravitational force, m is the mass of the body and g is the acceleration due to gravity.

The expression for the centripetal force is given by

$F_{C} = m \frac{v^{2}}{R}$

Here m is the mass of the body, v is the tangential velocity, R is the radius of the curvature and $F_{c}$ is the centripetal force.

Calculating the expression for the tension

(a)

There are only two forces acting on the mass $m_{2}$ . One of them is gravitational force in downward direction and the other one is tension in the upward direction. So gravitational force is balanced by tension in the string.

. $T = m_{2} g$ ..... (1)

Therefore, the expression for the tension in the string is $T = m_{2} g$ .

Calculating the radial force acting on the puck

(b)

The radial force acting on the puck to move the mass $m_{1}$ in a circular path is equal to the tension in the string.

$F_{R} = T$

From equation (1), we can write

$F_{R} = m_{2} g$

Therefore the radial force acting on the puck is $F_{R} = m_{2} g$ .

Calculate the speed of the puck

(c)

The tension in the string of mass provides the centripetal force to mass $m_{2}$ to rotate in circular path. So cetripetal force will be equal to the tesnion in the string.

$\begin{array}{l} m_{R}^{v} = T \\ x = m g \\ v = \sqrt{[] m]_{g} g} m \end{array}$

Therefore, the speed of the puck is $v = \sqrt{\frac{m_{2} g R}{m_{1}}}$ .

Explanation for (d)

(d)

If the mass $m_{2}$ will increase then the tension in string increases and mass $m_{2}$ will start moving in downward direction. It will also cause the mass $m_{1}$ to rotate in smaller circular radius.

: Explanation for (e)

(e)

If the mass $m_{2}$ will decrease then the tension in string decreases and mass $m_{2}$ will start moving in upward direction. It will also cause the mass $m_{1}$ to rotate in larger circular radius.