Question

A 0.06 kg ball is attached to the end of a 0.30 m long string. The ball is spun around a circle with a period of 0.90 s. What is the tension force in the string? (need this asap)

Answer 1

This question involves the concepts of tension force, centripetal force, and time period.

The tension force in the string is "0.88 N".

STEP 1:

We will find the angular speed of the ball:

$\omega = \frac{2\pi}{T}$

where,

ω = angualar speed = ?

T = time period = 0.9 s

$\omega = \frac{2\pi}{0.9\ s}$

ω = 6.98 rad/s

STEP 2:

We will find the linear speed of the ball:

v = rω

whre,

v = linear speed = ?

r = radius of circular path = length of string = 0.3 m

Therefore,

v = (0.3 m)(6.98 rad/s)

v = 2.09 m/s

STEP 3:

We will find the tension force in the string:

In order for the ball to stay on the circular path the tension force in the string must be equal to the centripetal force:

$Tension = Centripetal\ Force\\\\ T = \frac{mv^2}{r}=\frac{(0.06\ kg)(2.09\ m/s)^2}{0.3\ m}$

T = 0.88 N

Learn more about centripetal force here:

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