Question

In a lab, a 0.025 kg cork attached to a string is swung around in a horizontal circle. The time it takes for the cork to complete one revolution is timed. If the radius of the circle is 0.30 m and one revolution takes 1.25 s, find the speed of the cork and the centripetal force applied to it

Answer 1

Answer:

v=1.5081 m/s

$F_c=0,189\ Nw$

Explanation:

Uniform Circular Motion

The cork is performing a circular motion which we assume to be uniform (constant angular speed or angular acceleration zero)

The centripetal force applied to it is given by

$F_c=m\ a_c$

where m is the mass and $a_c$ is the centripetal acceleration. This acceleration appears since the tangent speed is constantly changing direction. If w is the angular speed and r is the radius of rotation

$a_c=w^2r$

The speed of the cork can be found with the formula

$v=wr$

We can compute w since we know the rotation period T=1.25 sec

$w=\frac{2\pi}{T}=\frac{2\pi}{1.25}=5.027\ rad/sec$

Now, since r=0.30 m, let's compute v

$v=wr=(5.027)(0.3)=1.5081\ m/s$

$a_c=5.027^2(0.3)=7.58\ m/sec^2$

And finally

$F_c=(0.025)\ (7.58)=0,189\ Nw$