Question

A 1.0 kg ball at the end of a 2.0 m string swings in a vertical plane. At its lowest point the ball is moving with a speed of 10 m/s. (a) What is its speed at the top of its path? (b) What is the tension in the string when the ball is at the bottom and at the top of its path?

Answer 1

Answer:

a) 4.65m/s

b) 59.8 N , 1.01125 N

Explanation:

a)

m = mass of the ball = 1 kg

r = length of the string = 2.0 m

h = height gained by the ball as it moves from lowest to topmost position = 2r = 2 x 2 = 4 m

v = speed at the lowest position = 10 m/s

v' = speed at the topmost position = ?

Using conservation of energy

Kinetic energy at topmost position + Potential energy at topmost position = Kinetic energy at lowest position

(0.5) m v'² + m g h = (0.5) m v²

(0.5) v'² + g h = (0.5) v²

(0.5) v'² + (9.8 x 4) = (0.5) (10)²

v' = 4.65m/s

b)

T' = Tension force in the string when the ball is at topmost position

T = Tension force in the string when the ball is at lowest position

At the topmost position:

force equation is given as

$mg + T' = \frac{m v'^{2}}{r}$

$(1)(9.8) + T' = \frac{(1) (4.65)^{2}}{2}$

T' = 1.01125 N

At the lowest position:

force equation is given as

$T - mg = \frac{m v^{2}}{r}$

$T - (1) (9.8) = \frac{(1) (10)^{2}}{2}$

T = 59.8 N