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)

5$\sqrt{2}$ m/s

Answer 2

The correct answer is (E): 5$\sqrt{2}$ m/s
To determine the velocity of the sphere when it reaches the top of the ring, we can use the principle of conservation of mechanical energy.
Given:

• Radius of the ring R =1.75m.
• Mass of the sphere 𝑚=m =1kg.
• Initial velocity at the bottom _v b_​=10m/s.
• Gravitational acceleration g =10m/s2.

At the bottom of the ring, the total mechanical energy is the sum of kinetic and potential energy. At the top, the sphere will have both kinetic and potential energy:
$E_{bottom}=\frac{1}{2}mv^2_b$
$E_{top}=\frac{1}{2}^2_t+mg(2R)$
Equating the energies (since mechanical energy is conserved):
$\frac{1}{2}mv^2_b=\frac{1}{2}mv^2_t+mg(2g)$
$\frac{1}{2}(1)(10)^2=\frac{1}{2}(1)v^2_t+(1)(10)(2\times1.75)$
$50=\frac{1}{2}v^2_t+35$
$v_t=5\sqrt2m/s$
Thus, the velocity of the sphere when it reaches the top of the ring is 5$\sqrt{2}$ m/s