A small ball of mass m=0.60 kg hangs from a massless string of length l=0.9 m. The ball travels in a vertical circle and its speed at the bottom is v0=6.0 m/s (see figure). Neglect all friction and air drag, and use g=10m/s^2 for the gravitational acceleration. The ball is so small that we can approximate it as a point.
(a) Find the speed of the ball (in m/s) when the string is at £\=50¢X
v(£\=50¢X)=
(b) What is the tension in the string (in Newton) when it is at £\=50¢X?
T(£\=50¢X)=
(c) The string of the pendulum is cut when it is at £\=50¢X. First, we want to neglect all air drag during the trajectory of the ball. What is the maximal height h (in meters) the ball reaches above its point of release? What time tup (in s) does it take the ball to reach the highest point from the instant the string is cut? What time tdn (in s) does it take the ball to go from the highest point back to the altitude it was released from the string?
h(£\=50¢X)=
tup(£\=50¢X)=
tdn(£\=50¢X)=
To solve this problem, we can use the conservation of energy and the principles of circular motion.
(a) To find the speed of the ball when the string is at £=\50¢X, we can equate the gravitational potential energy at the top with the kinetic energy at the bottom.
At the top of the circle, all of the ball's energy is in the form of gravitational potential energy:
mgh = 0.6 kg * 10 m/s^2 * (0.9 m - 0.9 cos(50°))
At the bottom of the circle, the energy is in the form of kinetic energy:
0.5mv^2 = 0.5 * 0.6 kg * (6.0 m/s)^2
Using these equations, we can solve for v, the speed of the ball at £=\50¢X.
(b) To find the tension in the string at £=\50¢X, we can consider the forces acting on the ball at that position. The tension in the string provides the centripetal force required to keep the ball moving in a circle. The gravitational force provides the weight of the ball at that position.
Tension + Weight = Centripetal force
T + mg = m(v^2 / r)
Using the value of v obtained in part (a) and the given radius of the circle, we can solve for T, the tension in the string at £=\50¢X.
(c) When the string is cut, the ball will follow a projectile motion trajectory. Neglecting air drag, we can use the conservation of mechanical energy to determine the ball's maximum height h above its point of release.
At the point of release, the energy is in the form of kinetic energy:
0.5mv0^2 = 0.5 * 0.6 kg * (6.0 m/s)^2
At the maximum height, the energy is in the form of gravitational potential energy:
mgh = 0.6 kg * 10 m/s^2 * h
Using these equations, we can solve for h, the maximum height reached by the ball.
To find the times tup and tdn, we can use the symmetry of the problem. The time it takes for the ball to reach the highest point from the instant the string is cut is equal to the time it takes for the ball to fall from the highest point back to the altitude it was released from the string.
We can use the kinematic equation for vertical motion:
h = 0.5 g t^2
Using this equation, we can solve for tup and tdn, the times it takes for the ball to reach the highest point and return to the same altitude, respectively.