Question

A hoop of radius $2\, m$ weighs $100 \,kg$. It rolls along a horizontal floor so that its centre of mass has a speed of $20 \,cm s^{-1}$ How much work has to be done to stop it?

• A. $2\,J$
• B. $4\,J$
• C. $6\,J$
• D. $8\,J$

Answer 1

The Correct Option is B

To find out how much work needs to be done to stop the hoop, we need to calculate its total kinetic energy and then determine how much of that energy needs to be removed to bring the hoop to rest.

The hoop has both translational and rotational kinetic energy since it is rolling. The formulas for these are:

• Translational kinetic energy: KE_{\text{trans}} = \frac{1}{2} m v^2
• Rotational kinetic energy: KE_{\text{rot}} = \frac{1}{2} I \omega^2

For a hoop, the moment of inertia (I) is m r^2 where m is the mass, and r is the radius. Also, the angular velocity (\omega) is related to the linear velocity (v) by \omega = \frac{v}{r}.

Given data:

• Mass of the hoop, m = 100 \, \text{kg}
• Radius of the hoop, r = 2 \, \text{m}
• Speed of the center of mass, v = 20 \, \text{cm/s} = 0.2 \, \text{m/s}

First, calculate the translational kinetic energy:

• KE_{\text{trans}} = \frac{1}{2} \times 100 \, \text{kg} \times (0.2 \, \text{m/s})^2 = 2 \, \text{J}

Next, calculate the rotational kinetic energy:

• I = 100 \, \text{kg} \times (2 \, \text{m})^2 = 400 \, \text{kg} \cdot \text{m}^2
• \omega = \frac{0.2 \, \text{m/s}}{2 \, \text{m}} = 0.1 \, \text{rad/s}
• KE_{\text{rot}} = \frac{1}{2} \times 400 \, \text{kg} \cdot \text{m}^2 \times (0.1 \, \text{rad/s})^2 = 2 \, \text{J}

Total kinetic energy of the hoop:

• KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = 2 \, \text{J} + 2 \, \text{J} = 4 \, \text{J}

The work done to stop the hoop must be equal to the total kinetic energy:

• The work needed = 4 \, \text{J}

Thus, the correct answer is 4 \, \text{J}.