Answer
Verified
444.3k+ views
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.
Recently Updated Pages
A particle is undergoing a horizontal circle of radius class 11 physics CBSE
A particle is thrown vertically upwards with a velocity class 11 physics CBSE
A particle is rotated in a vertical circle by connecting class 11 physics CBSE
A particle is projected with a velocity v such that class 11 physics CBSE
A particle is projected with a velocity u making an class 11 physics CBSE
A particle is projected vertically upwards and it reaches class 11 physics CBSE
Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
In the tincture of iodine which is solute and solv class 11 chemistry CBSE
On which part of the tongue most of the taste gets class 11 biology CBSE
State and prove Bernoullis theorem class 11 physics CBSE
Who is the leader of the Lok Sabha A Chief Minister class 11 social science CBSE