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What is the minimum velocity with which a body of mass m m m must enter a vertical loop of a radius R R R so that it can complete the loop?
hard
Gravitation
2016
physics
Explanation To solve this problem, we need to determine the minimum velocity required for a body to complete a vertical loop of radius R . R. \\
R . Consider the forces acting on the body at the top of the loop. At this point, the gravitational force and the centripetal force must be balanced. \\
The gravitational force acting on the body is m g , mg, m g , where m m m is the mass of the body and g g g
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is the acceleration due to gravity.
The centripetal force required to keep the body moving in a circle is
m v 2 R , \frac{mv^2}{R}, R m v 2 , where
is the velocity of the body at the top of the loop.
For the body to just complete the loop, the centripetal force must be equal to the gravitational force at the top of the loop:
m v 2 R = m g \\
\frac{mv^2}{R} = mg \\
R m v 2 = m g Simplifying, we get:
v 2 = g R \\
v^2 = gR \\
v 2 = g R Thus, the minimum velocity at the top of the loop is
v = g R . v = \sqrt{gR}. \\
v = g R . However, to find the minimum velocity at the bottom of the loop, we need to consider the conservation of energy.
At the bottom of the loop, the body has kinetic energy and potential energy. At the top of the loop, it has only potential energy.
Let
be the velocity at the bottom of the loop. The total mechanical energy at the bottom is:
1 2 m v 0 2 \\
\frac{1}{2}mv_0^2 \\
2 1 m v 0 2 At the top of the loop, the total mechanical energy is:
1 2 m v 2 + 2 m g R \\
\frac{1}{2}mv^2 + 2mgR \\
2 1 m v 2 + 2 m g R Using conservation of energy, we equate the energies at the bottom and top of the loop:
1 2 m v 0 2 = 1 2 m v 2 + 2 m g R \\
\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + 2mgR \\
2 1 m v 0 2 = 2 1 m v 2 + 2 m g R Substitute
into the equation:
1 2 m v 0 2 = 1 2 m ( g R ) + 2 m g R \\
\frac{1}{2}mv_0^2 = \frac{1}{2}m(gR) + 2mgR \\
2 1 m v 0 2 = 2 1 m ( g R ) + 2 m g R Simplify:
1 2 m v 0 2 = 1 2 m g R + 2 m g R 1 2 m v 0 2 = 5 2 m g R \\
\frac{1}{2}mv_0^2 = \frac{1}{2}mgR + 2mgR \\
\frac{1}{2}mv_0^2 = \frac{5}{2}mgR \\
2 1 m v 0 2 = 2 1 m g R + 2 m g R 2 1 m v 0 2 = 2 5 m g R Solve for
v 0 2 : v 0 2 = 5 g R v_0^2: \\
v_0^2 = 5gR \\
v 0 2 : v 0 2 = 5 g R Thus, the minimum velocity at the bottom of the loop is
v 0 = 5 g R . v_0 = \sqrt{5gR}. \\
v 0 = 5 g R . Therefore, the correct answer is Option 2: