
A stone of mass \[0.25kg\] tied to the end of a string is whirled round in a circle of radius \[1.5m\] with a speed of \[40rev/\min \] 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 \[200N\] ?
Answer
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Hint: We start by gathering the information given to us and finding the tension in the string using the formula given below. Then by using the formula for maximum tension, we find the value of maximum velocity.
Formulas used: Tension in the string (radius of the string) is given by the formula, \[T = m{\omega ^2}r\]
The maximum tension in the radius of the circle is given by the formula, \[{T_{\max }} = \dfrac{{m{v_{\max }}^2}}{r}\]
The angular velocity of the stone is given by the formula, \[\omega = 2\pi n\]
Where, \[m\] is the mass of the stone
\[\omega \] is the angular velocity
\[r\] is the radius of the circle
\[{v_{\max }}\] is the maximum velocity that the string can withstand
Complete step by step solution:
The following information is given,
Mass of the stone is, \[m = 0.25kg\]
Radius of the circular path is, \[r = 1.5m\]
Number of revolutions per second, \[n = \dfrac{{40revs}}{{60s}} = \dfrac{2}{3}rps\]
Maximum value of tension is, \[{T_{\max }} = 200N\]
We find the value of tension using the formula, \[T = m{\omega ^2}r\]
We get, \[T = 0.25 \times {\left( {2\pi \times \dfrac{2}{3}} \right)^2}1.5 = 6.57N\]
It is given that the maximum value of tension is \[{T_{\max }} = 200N\]
Now we use the formula, \[{T_{\max }} = \dfrac{{m{v_{\max }}^2}}{r}\] to find the maximum value of velocity
\[{v_{\max }} = \sqrt {\dfrac{{r{T_{\max }}}}{m}} = \sqrt {\dfrac{{1.5 \times 200}}{{0.25}}} = 34.64m/s\]
The maximum velocity is, \[34.64m/s\]
Note:
Tension is the axial pulling force by means of a string or cable. The kinetic energy at the lowest point during the motion where the tension will be the greatest. circular motion is when a body is moving by the circumference of a circle or when the vibration of a body is in a circle. Maximum velocity of a circular motion is when the body is at the bottom of the circle, because this point has the least potential energy.
Formulas used: Tension in the string (radius of the string) is given by the formula, \[T = m{\omega ^2}r\]
The maximum tension in the radius of the circle is given by the formula, \[{T_{\max }} = \dfrac{{m{v_{\max }}^2}}{r}\]
The angular velocity of the stone is given by the formula, \[\omega = 2\pi n\]
Where, \[m\] is the mass of the stone
\[\omega \] is the angular velocity
\[r\] is the radius of the circle
\[{v_{\max }}\] is the maximum velocity that the string can withstand
Complete step by step solution:
The following information is given,
Mass of the stone is, \[m = 0.25kg\]
Radius of the circular path is, \[r = 1.5m\]
Number of revolutions per second, \[n = \dfrac{{40revs}}{{60s}} = \dfrac{2}{3}rps\]
Maximum value of tension is, \[{T_{\max }} = 200N\]
We find the value of tension using the formula, \[T = m{\omega ^2}r\]
We get, \[T = 0.25 \times {\left( {2\pi \times \dfrac{2}{3}} \right)^2}1.5 = 6.57N\]
It is given that the maximum value of tension is \[{T_{\max }} = 200N\]
Now we use the formula, \[{T_{\max }} = \dfrac{{m{v_{\max }}^2}}{r}\] to find the maximum value of velocity
\[{v_{\max }} = \sqrt {\dfrac{{r{T_{\max }}}}{m}} = \sqrt {\dfrac{{1.5 \times 200}}{{0.25}}} = 34.64m/s\]
The maximum velocity is, \[34.64m/s\]
Note:
Tension is the axial pulling force by means of a string or cable. The kinetic energy at the lowest point during the motion where the tension will be the greatest. circular motion is when a body is moving by the circumference of a circle or when the vibration of a body is in a circle. Maximum velocity of a circular motion is when the body is at the bottom of the circle, because this point has the least potential energy.
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