
A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is:
A. \[\sqrt {gl} \]
B. \[\sqrt {2gl} \]
C. \[\sqrt {3gl} \]
D. \[\sqrt {5gl} \]
Answer
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Hint: It must satisfy the constraints of centripetal force to remain in a circle and must satisfy demands of conservation of energy as gravitational potential energy is converted to kinetic energy when the mass moves downward. At the lowest point in the vertical circular motion, the minimum velocity of the particle to complete the circular motion is \[\sqrt {5gl} \].
Complete step by step answer:
We know that the minimum velocity of the body at the bottom position to complete one complete vertical revolution is \[\sqrt {5gl} \]. We can use the law of conservation of energy to calculate the minimum velocity at the horizontal position. As the body moves from the bottom position to the horizontal position, the loss in the kinetic energy is converted into the potential energy.
Therefore,
\[\dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2 + mgl\]
Here, m is the mass of the particle, \[{v_1}\] is the velocity at the bottom position and \[{v_2}\] is the velocity at the horizontal position, g is the acceleration due to gravity and l is the length of the string.
Substituting \[\sqrt {5gl} \] for \[{v_1}\], we get,
\[\dfrac{1}{2}m\left( {5gl} \right) = \dfrac{1}{2}mv_2^2 + mgl\]
\[ \Rightarrow \dfrac{5}{2}gl = \dfrac{1}{2}v_2^2 + gl\]
\[ \Rightarrow \dfrac{5}{2}gl - gl = \dfrac{1}{2}v_2^2\]
\[ \Rightarrow \dfrac{3}{2}gl = \dfrac{1}{2}v_2^2\]
\[ \therefore {v_2} = \sqrt {3gl} \]
So, the correct answer is option C.
Note: In this motion the centripetal force, the force that points towards the center of the circle, does not remain constant since the velocity of the particle changes at every position. At the bottom of the circle gravity is pointing in the opposite direction to the tension. In the circular motion, the distance of the particle from the centre always remains constant.
Complete step by step answer:
We know that the minimum velocity of the body at the bottom position to complete one complete vertical revolution is \[\sqrt {5gl} \]. We can use the law of conservation of energy to calculate the minimum velocity at the horizontal position. As the body moves from the bottom position to the horizontal position, the loss in the kinetic energy is converted into the potential energy.
Therefore,
\[\dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2 + mgl\]
Here, m is the mass of the particle, \[{v_1}\] is the velocity at the bottom position and \[{v_2}\] is the velocity at the horizontal position, g is the acceleration due to gravity and l is the length of the string.
Substituting \[\sqrt {5gl} \] for \[{v_1}\], we get,
\[\dfrac{1}{2}m\left( {5gl} \right) = \dfrac{1}{2}mv_2^2 + mgl\]
\[ \Rightarrow \dfrac{5}{2}gl = \dfrac{1}{2}v_2^2 + gl\]
\[ \Rightarrow \dfrac{5}{2}gl - gl = \dfrac{1}{2}v_2^2\]
\[ \Rightarrow \dfrac{3}{2}gl = \dfrac{1}{2}v_2^2\]
\[ \therefore {v_2} = \sqrt {3gl} \]
So, the correct answer is option C.
Note: In this motion the centripetal force, the force that points towards the center of the circle, does not remain constant since the velocity of the particle changes at every position. At the bottom of the circle gravity is pointing in the opposite direction to the tension. In the circular motion, the distance of the particle from the centre always remains constant.
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