Question

A ball with a mass of 270 g is tied to a light string that has a
length of 2.40 m. The end of the string is tied to a hook, and the
ball hangs motionless below the hook. Keeping the string taut, you
move the ball back and up until it is a vertical distance of 1.16 m
above its equilibrium point. You then release the ball from rest,
and it oscillates back and forth, pendulum style. As usual, we will
neglect air resistance. Use *g* = 9.80 m/s^{2}.

(a) What is the highest speed the ball achieves in its
subsequent motion?

m/s

(b) What is the maximum height the ball reaches in its subsequent
motion (measured from its equilibrium position)?

m

(c) When the ball passes through the equilibrium point for the
first time, what is the magnitude of the tension in the
string?

N

Answer #1

given

m=0.27 kg

r = 2.4 m

a) let the equlibrium point PE = 0

and h = 1.16 m

ball will be at max vvelocity the it is passing through its equilibrium point

from energy conservation we can write

KEi + PEi = KEf + PEf

0 + mgh = 1/2 mv2 + 0

mx9.8 x 1.16 = 1/2 m v^{2}

vmax = 4.76 m/s

b)

as air resistance is negligible

so from energy conservation ball reaches in its subsequent motion

same height

so h = 1.16 m

c) at equilibrium point

T= mg +mv^{2}/R

T = (0.27x9.8) + 0.27 x 4.76^{2} / 2.4

T = 5.2 N

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