Question

A ball of mass 0.40 kg is attached to a light rope of length 1.0 m. The ball is moving in a vertical circle, as shown in the figure, with Earth's gravitational force as the only force external to the ball string system. The speed of the ball at point is 5.4 The acceleration of the ball at point is most nearly

Answer 1

1.0 m A ball of mass 0.40 kg is attached to a light rope of length 1.0 m. The ball is moving in a vertical circle, as shown in the figure, with Earth's gravitational force as the only force external to the ball-string system. The speed of the ball at point P is 5.4 m/s The acceleration of the ball at point P is most nearly 9.8 m/s/s 31

Explanation:

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Answer 2

Given: mass of a ball = 0.40 kg

           length of rope = 1 m

           speed of the ball at a point = 5.4 m/s

To find: Acceleration of the ball at the point

Solution:

• In a uniform circular motion, tangential acceleration will always be zero. Only centripetal acceleration will be there.

• In a circular motion, total acceleration is composed of tangential acceleration and radial acceleration.

• Acceleration in a circular motion is calculated by the ratio of the product of the square of velocity and radius of the circular path.
• The acceleration which is responsible for the change in the path of the ball in a circular motion and is directed towards the centre hence is called centripetal force.

Acceleration = v²/r ( v is the velocity, r is the radius of the path)

= (5.4)²/1 = 29.16 ms⁻²

Therefore, the acceleration of the ball will be 29.16 ms⁻²