a small block with mass 0.390 kg is attached to a string passing through a hole in a frictionless, horizontal surface (figure 1). the block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. the string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. the breaking strength of the string is 30.0 n.

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ShubhamSrivastava ShubhamSrivastava · Answer 1 · 2022-12-13T07:12:36+01:00

The radius of the circle when stringe break is 6.93×10^-2.

The radius is the distance from the centre of a circle.

When finding the radius of the string at the point it breaks, the tangential

velocity is assumed to be constant.

The radius when the string breaks is 6.9.3×10^-3m

Reasons:

The mass of the small block, m = 0.130 kg

Initial radius of the circle of rotation = 0.800 m

Tangential velocity, v = 4.00 m/s

The radius of the path of rotation is reduced as the string is pulled

Breaking strength of the string = 30.0 N

Required:

The radius of the circle when the string brakes

Solution:

centripetal force =m.v^2/r

Where;

r = The radius of the circle of rotation

When the string brakes, w have;

Centripetal force = Breaking strength of the string = 30.0 N

Which gives;

r=m.v^2/centrigual force =6.93×10^-2

The radius of the circle when, the string breaks r = =6.93×10^-2.

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