A ball of mass m=4.6, at one end of a string of length L=6.6, 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:
at the top:
mv^2/r=mg
v^2=rg
KE at top: 1/2 m v^2=1/2 m rg
PE at top mag(2r)
KE at bottome+1/2 m rg+mg(2r)
1/2 mvb^2=1/2 mrg+2rmg
vb^2=5rg
veloecitybaottom=sqrt(5rg
To find the speed of the ball at the bottom of the circle, we need to consider the conservation of mechanical energy.
At the top of the circle, when the string is about to go slack, the ball is at its highest point of the circular motion. Let's call this point A. At point A, the ball has only potential energy (since it is not moving). The formula for potential energy is:
Potential Energy (PE) = mgh
Where m is the mass of the ball, g is the acceleration due to gravity, and h is the height from the lowest point of the circle to the highest point.
At the bottom of the circle, let's call this point B, the ball has both potential energy and kinetic energy. The formula for kinetic energy is:
Kinetic Energy (KE) = 1/2 mv^2
Where m is the mass of the ball and v is the velocity/speed of the ball.
Since we are assuming mechanical energy is conserved, the total mechanical energy at point A (top) is equal to the total mechanical energy at point B (bottom):
PE(A) = KE(B)
mgh(A) = 1/2 mv^2(B)
Since the mass of the ball cancels out, we're left with:
gh(A) = 1/2 v^2(B)
Now, let's solve for v(B), the speed of the ball at the bottom of the circle:
v(B) = sqrt(2gh(A))
Substituting the given values:
v(B) = sqrt(2 * 9.8 m/s^2 * 6.6 m)
v(B) ≈ 10.63 m/s
Therefore, the speed of the ball at the bottom of the circle is approximately 10.63 m/s.