A small sphere of radius r is placed on a con | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) When a small sphere of radius $r$ rolls without slipping on a concave surface of radius $R$, the effective radius of the path of the center of mas

Vedclass · Accepted Answer

$2 \pi \sqrt{\frac{R}{g}}$

Vedclass · Answer

(A) When a small sphere of radius $r$ rolls without slipping on a concave surface of radius $R$, the effective radius of the path of the center of mass is $(R - r)$.The torque about the point of contact is $	au = -mg(R-r)\sin	heta \approx -mg(R-r)	heta$.The moment of inertia of the sphere about the point of contact is $I = I_{cm} + mr^2 = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2$.The equation of motion is $	au = I\alpha$, so $-mg(R-r)	heta = \frac{7}{5}mr^2 \frac{d^2	heta}{dt^2}$.This gives $\frac{d^2	heta}{dt^2} = -\frac{5g(R-r)}{7r^2}	heta$.However, if the sphere is assumed to slide without friction, the motion is equivalent to a simple pendulum of length $l = (R-r)$.Given $r The time period is $T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{R}{g}}$.

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

Similar Questions

$A$ simple pendulum of length $1 \,m$ is freely suspended from the ceiling of an elevator. The time period of small oscillations as the elevator moves up with an acceleration of $2 \,m/s^2$ is (use $g=10 \,m/s^2$).

$A$ pendulum suspended from the ceiling of a train has a period $T$,when the train is at rest. When the train is accelerating with a uniform acceleration $a$,the period of oscillation will

$A$ small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6 \ m$. The period of oscillation of the sphere is (acceleration due to gravity $g = 10 \ m/s^2$) (in $\pi \ s$)

In a simple pendulum,the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle $\theta$ with the vertical.

$A$ simple pendulum keeps correct time at $0^{\circ}C$. At $25^{\circ}C$,it loses $12.5 \, s$ in a day. What is the coefficient of linear expansion of the metal of the pendulum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module