The period of a disk orf radius 10 cm executing small oscillation about a piviot at its rim is measured to be 0.784 s. Find the value of g, the acceleration due to gravity

The earth rotates about its own axis, then the value of acceleration due to gravity is

A wheel of radius 15 cm is rotating about its axis at angluar speed of 20 rad//s . Find the linear speed and total acceleration of a point on the rim.

The angular speed of rotation of the earth is omega=7.27xx10^(-5)rads^(-1) and its radius is R=6.37xx10^(6)m. Calculate the acceleration of a man standing at a place at 40^(@) latitude. [cos40^(@)=10.77] If the earth suddenly stops rotating, the acceleration due to gravity on its surface will become g_(0)=9.82ms^(-2). Find the effective value of acceleration due to gravity (g) at 40^(@) latitude taking into account the rotation of the earth.

A disc of radius R is pivoted at its rim. The period for small oscillations about an axis perpendicular to the plane of disc is

A disc of radius 10 cm is rotating about its axis at an angular speed of 20 rad/s. Find the linear speed of a. a point on the rim, b. the middle point of a radius.

A satellite of mass m is orbiting the earth of radius R at) a height h from its surface. The total energy of the satellite in terms of g_(0) the value of acceleration due to gravity at the earth 's surface, is

The period of oscillation of a simple pendulum is T = 2 pi sqrt((L)/(g)) .L is about 10 cm and is known to 1mm accuracy . The period of oscillation is about 0.5 s . The time of 100 oscillation is measured with a wrist watch of 1 s resolution . What is the accuracy in the determination of g ?