A ball of mass m=8.5, at one end of a string of length L=3.6, rotates in a vertical circle just fast enough to prevent the string from going slack at the top of the circle. Assuming mechanical energy is conserved, the speed of the ball at the bottom of the circle is:
A ball of mass m, at one end of a string of length L, rotates in a vertical circle just fast enough to prevent the string from going slack at the top of the circle. Assuming mechanical energy is conserved, the speed of the ball at the bottom of the circle is:
To find the speed of the ball at the bottom of the circle, we can start by using the conservation of mechanical energy. At the top of the circle, the ball has gravitational potential energy and kinetic energy, and at the bottom of the circle, all of its energy is in the form of kinetic energy.
Let's break down the conservation of mechanical energy equation:
1. Gravitational potential energy (U_g) at the top of the circle:
U_g = mgh
Here, m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the ball at the top of the circle. In this case, h is the length of the string (L) since the string is vertical.
U_g = mgL
2. Kinetic energy (K) at the top of the circle:
K = 0.5mv^2
Here, v is the velocity of the ball at the top of the circle.
Now, since mechanical energy is conserved, the total energy at the top of the circle (U_g + K) should be equal to the total energy at the bottom of the circle (K):
mgL + 0.5mv^2 = 0.5mv_b^2
Here, v_b is the velocity of the ball at the bottom of the circle, which is what we're trying to find.
To solve for v_b, we need to rearrange the equation:
0.5mv^2 - 0.5mv_b^2 = mgL
0.5m(v^2 - v_b^2) = mgL
(v^2 - v_b^2) = 2gL
v^2 - v_b^2 = 2gL
Now, we can solve this equation for v_b:
v_b = sqrt(v^2 - 2gL)
Substituting in the given values of m, L, and g:
v_b = sqrt(v^2 - 2 * 9.8 * 3.6)
v_b = sqrt(v^2 - 70.56)
Therefore, the speed of the ball at the bottom of the circle is given by sqrt(v^2 - 70.56).