A small sphere of radius r is placed as a concave surface of radius of curvature R a little away from the centre. When the sphere is released, it oscillates. Assuming the oscillation to be simple harmonic motion, and r

Question

A small sphere of radius r is placed as a concave surface of radius of curvature R a little away from the centre. When the sphere is released, it oscillates. Assuming the oscillation to be simple harmonic motion, and r< < R then the time period is (A) 2 π √(R/g) (B) 2 π √(3 R/2 g) (C) 2 π √(2 R/3 g) (D) 2 π √(R/2 g)

Tardigrade Admin · Accepted Answer

The motion of sphere is as shown in figure <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a09cd6ed55796d3f9d3bbbd29b403331-.png" /> As, the sphere has simple harmonic motion on a curved surface, which is equivalent to the motion of pendulum. So, its time period is T=2 π √(I/g) Here, l=R ∴ T=2 π √R / g

Q. A small sphere of radius $r$ is placed as a concave surface of radius of curvature $R$ a little away from the centre. When the sphere is released, it oscillates. Assuming the oscillation to be simple harmonic motion, and $r << R$ then the time period is

$2 π \frac{R}{g}$

$2 π \frac{3 R}{2 g}$

$2 π \frac{2 R}{3 g}$

$2 π \frac{R}{2 g}$

The motion of sphere is as shown in figure

As, the sphere has simple harmonic motion on a curved surface,
which is equivalent to the motion of pendulum.
So, its time period is
$T = 2 π \frac{I}{g}$
Here, $l = R$
$∴ T = 2 π R / g$

Q. A small sphere of radius r is placed as a concave surface of radius of curvature R a little away from the centre. When the sphere is released, it oscillates. Assuming the oscillation to be simple harmonic motion, and r<<R then the time period is

2πgR​​

2π2g3R​​

2π3g2R​​

2π2gR​​