Question

Suppose David puts a 0.60-kg rock into a sling of length 1.5 m and begins whirling the rock in a nearly horizontal circle, accelerating it from rest to a rate of 75 rpm after 5.0 s. What is the torque required to achieve this feat, and where does the torque come from?

Short answer

The torque required to whirl the rock is \(2.1\;{\rm{m}} \cdot {\rm{N}}\). This torque is provided by the arm muscles of David with the help of which he whirls the rock in a horizontal circle.

Step-by-step solution

Newton’s second law for rotation

According to Newton’s second law for rotation, when a rigid object rotates about a fixed axis, thetorque \(\left( \tau \right)\)acting on the object is equal to the product of the moment of inertia \(\left( I \right)\) and the angular acceleration \(\left( \alpha \right)\)of the object, i.e.,

\(\tau = I\alpha \)

In this problem,the torque required to achieve the rotational motion is equal to the product of the moment of inertia and angular acceleration of the rock.

Given information

The mass of the rock is m = 0.60 kg.

The length of the string is 1.5 m.

The radius of the horizontal circle formed by the whirling of the rock is equal to the length of the string, i.e., \(r = 1.5\;{\rm{m}}\).

The initial angular velocity of the rock is\({\omega _0} = 0\;{\rm{rad/s}}\).

The final angular velocity of the rock is:

\(\begin{aligned}{c}\omega = 75\;{\rm{rpm}}\\ = \left( {\frac{{75\;{\rm{rev}}}}{{1\;\min }}} \right) \times \left( {\frac{{\;1\;\min }}{{60\;{\rm{s}}}}} \right) \times \left( {\frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}}} \right)\\ = 7.85\;{\rm{rad/s}}\end{aligned}\)

The total time taken is, \(t = 5.0\;{\rm{s}}\)

Determination of the angular acceleration of the whirling rock

Let\(\alpha \)be the angular acceleration of the whirling rock.

Using the kinematic equation of motion for the constant angular acceleration, you will get:

\(\begin{aligned}{c}\omega = {\omega _0} + \alpha t\\\alpha = \frac{{\omega - {\omega _0}}}{t}\\ = \frac{{\left( {7.85\;{\rm{rad/s}}} \right) - \left( {0\;{\rm{m/s}}} \right)}}{{\left( {5.0\;{\rm{s}}} \right)}}\\ = 1.57\;{\rm{m/s}}\end{aligned}\) ... (i)

Thus, the angular acceleration of the whirling rock is 1.57 m/s.

Determination of the torque

According to Newton’s second law for rotation, the torque required to whirl the rock is given as:

\(\begin{aligned}{c}\tau = I\alpha \\ = m{r^2}\alpha \\ = \left( {0.60\;{\rm{kg}}} \right){\left( {1.5\;{\rm{m}}} \right)^2}\left( {1.57\;{\rm{rad/}}{{\rm{s}}^2}} \right)\\ = 2.1\;{\rm{m}} \cdot {\rm{N}}\end{aligned}\)

This torque is provided by the muscles of the arm of David with the help of which he whirls the rock in a horizontal circle.