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 α=50∘.

v=(α=50∘)=

(b) What is the tension in the string (in Newton) when it is at α=50∘?

T=(α=50∘)=

(c) The string of the pendulum is cut when it is at α=50∘. 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=(α=50∘)=
tup=(α=50∘)=
tdn=(α=50∘)=

Come on bro, study a little bit

To solve these problems, we can use conservation of energy principles and apply the equations for circular motion.

(a) To find the speed of the ball when the string is at α = 50°, we can use conservation of energy. At the bottom, the ball has kinetic energy (K.E.) and gravitational potential energy (G.P.E.). At α = 50°, all the energy is in the form of K.E. (since the ball is at the highest point of the trajectory), so we can equate the two types of energy:

K.E. = G.P.E.

At the bottom, the kinetic energy is given by

K.E. = (1/2) * m * v0^2 = (1/2) * 0.60 kg * (6.0 m/s)^2

The gravitational potential energy at height h is given by

G.P.E. = m * g * h

Since the ball is at the highest point, h is the maximum height above its point of release, which we'll find later.

Setting K.E. = G.P.E., we have

(1/2) * m * v0^2 = m * g * h

Now we can solve for v (α=50°):

v (α=50°) = √(2gh)

(b) To find the tension in the string when α = 50°, we need to consider the forces acting on the ball at that point. The tension in the string provides the necessary centripetal force to keep the ball moving in a circle. At the highest point (α = 50°), the tension and gravitational force must add up vectorially to provide the centripetal force.

The net force on the ball is given by:

Net force = T - m * g = m * v^2 / (l * cos(α))

where T is the tension in the string, m is the mass of the ball, g is the acceleration due to gravity, l is the length of the string, and α is the angle the string makes with the vertical.

Since we need the tension when α = 50°, we can plug in the known values (m, g, l, α = 50°) and solve for T.

T (α = 50°) = m * (g + v^2 / (l * cos(α)))

(c) To find the maximal height h, we can again use conservation of energy principles. Initially, the ball has both kinetic energy and gravitational potential energy. When the string is cut, all the energy is in the form of G.P.E at the highest point. Using the same equation as in part (a), we can solve for h:

(1/2) * m * v0^2 = m * g * h

Solving for h, we get:

h = (v0^2) / (2 * g)

To find the time it takes the ball to reach the highest point (tup) and the time it takes to go from the highest point back to the altitude it was released from the string (tdn), we need to consider the total time taken for the entire motion of the ball.

The time it takes to reach the highest point (tup) is equal to the time taken for half the circular motion (since the total motion consists of going up to the highest point and then back down). We can use the formula for the time period of a circular motion to calculate tup:

tup = T = (2π * l) / v0

The time it takes to go from the highest point back to the altitude it was released from the string (tdn) is also equal to the time taken for half the circular motion. Therefore, tdn will also be equal to tup:

tdn = tup = (2π * l) / v0

Now we can substitute the known values into the above equations to find the solutions for (a), (b), (c).