Question

A ring of radius 1.75 m stands vertically. A small sphere of mass 1kg rolls on the inside of this ring without slipping. If it has a velocity of 10m/s at the bottom of the ring,then its velocity when it reaches the top is:(Take g=10 m/s²)

• A. 3\(\sqrt{2}\) m/s
• B. 2\(\sqrt{3}\) m/s
• C. 8\(\sqrt{2}\) m/s
• D. 2\(\sqrt{5}\) m/s
• E. 5\(\sqrt{2}\) m/s

Answer 1

Answer: (E)

Step 1: Total Energy at the Bottom

At the bottom of the ring:

Translational Kinetic Energy: \[ KE_{trans} = \frac{1}{2}mv^2 \]

Rotational Kinetic Energy:

For a rolling object without slipping, \(v = r\omega\).

For a solid sphere, moment of inertia \(I = \frac{2}{5}mr^2\).

\[ KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2}{5}mr^2 \times \left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2 \]

Total Kinetic Energy: \[ KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \]

Potential Energy: Considered zero at the bottom.

Step 2: Total Energy at the Top

At the top of the ring:

Height Change: \(h = 2r\)

Potential Energy: \[ PE = mgh = mg \times 2r \]

Kinetic Energy (with velocity \(v'\)): \[ KE_{total}' = \frac{7}{10}mv'^2 \]

Step 3: Conservation of Energy

Total energy at bottom = Total energy at top:

\[ \frac{7}{10}mv^2 = \frac{7}{10}mv'^2 + 2mgr \]

Mass \(m\) cancels out:

\[ \frac{7}{10}v^2 = \frac{7}{10}v'^2 + 2gr \]

Step 4: Substitute Values and Solve

Given:

• \(v = 10 \, \text{m/s}\)
• \(g = 10 \, \text{m/s}^2\)
• \(r = 1.75 \, \text{m}\)

\[ \frac{7}{10} \times (10)^2 = \frac{7}{10}v'^2 + 2 \times 10 \times 1.75 \] \[ 70 = 0.7v'^2 + 35 \] \[ 35 = 0.7v'^2 \] \[ v'^2 = \frac{35}{0.7} = 50 \] \[ v' = \sqrt{50} = 5\sqrt{2} \, \text{m/s} \]

Answer 2

1. Apply conservation of mechanical energy:

As the sphere rolls without slipping, mechanical energy (the sum of kinetic and potential energy) is conserved. The sphere's kinetic energy has both translational and rotational components.

2. Define variables:

• m = 1 kg (mass of sphere)
• r = 1.75 m (radius of ring)
• v_bottom = 10 m/s (velocity at bottom)
• v_top = ? (velocity at top)
• g = 10 m/s² (acceleration due to gravity)
• h = 2r = 3.5 m (change in height from bottom to top)

3. Set up the energy conservation equation:

\[KE_{bottom} + PE_{bottom} = KE_{top} + PE_{top}\]

Since the sphere is at its lowest point at the bottom, let \(PE_{bottom} = 0\). Then \(PE_{top} = mgh\).

The kinetic energy of the sphere has a translational and a rotational component. Since the sphere rolls without slipping, we can write the rotational kinetic energy in terms of v: 

\(KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{5}mv^2 \) So, total kinetic energy at any given point is: \( KE = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \).

\[\frac{7}{10}mv_{bottom}^2 + 0 = \frac{7}{10}mv_{top}^2 + mgh\]

4. Solve for \(v_{top}\):

\[\frac{7}{10}m(10^2) = \frac{7}{10}mv_{top}^2 + m(10)(3.5)\]

Divide through by m:

\[\frac{7}{10}(100) = \frac{7}{10}v_{top}^2 + 35\]

\[70 = \frac{7}{10}v_{top}^2 + 35\]

\[35 = \frac{7}{10}v_{top}^2\]

\[v_{top}^2 = 50\]

\[v_{top} = \sqrt{50} = 5\sqrt{2} \, m/s\]