Question

A stone of mass 0.25kg tied to the end of string is whirled round in a circle of radius 1.5m with a speed of 40 rev. /min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200N




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Answer 1

Answer:

Tension = 6.58 N

Maximum speed  = 34.6 m/s

Explanation:

Given:

• Mass of the stone = 0.25 kg
• Radius of the circular path = 1.5 m
• Frequency = 40 rev/min
• Maximum tension = 200 N

To Find:

• Tension in the string
• Maximum speed with which the stone can be whirled around if the maximum tension it can withstand is 200 N

Solution:

The tension produced in the string is given by the centripetal force which is given by,

T = m v²/r

T = m r ω² (∵ ω = v/r)

where m is the mass of stone,

r is the radius

ω is the angular velocity

First finding the angular velocity of the stone ω

ω = 2 π f

where f is the frequency,

Substitute the data,

ω = 2 × π × 40 = 80 π rad/min = 4.189 rad/s

Hence,

T = 0.25 × 1.5 × 4.189²

T = 6.58 N

Hence the tension produced in the string is 6.58 N

We know that maximum tension is given by,

$\sf T_{max}=\dfrac{m\:v_{max}^{2} }{r}$

Substitute the data,

$\sf 200=\dfrac{0.25\times v_{max}^{2} }{1.5}$

$\sf 200=0.167\times v_{max}^{2}$

$\sf v_{max}^{2} =1197.6$

$\sf v_{max}=\sqrt{1197.6}$

$\sf v_{max}=34.6\:m/s$

Hence the stone can be whirled with a maximum speed of 34.6 m/s.

Answer 2

Answer:

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