Question

A billiard ball (mass m = 0.150 kg) is attached to a light string that is 0.50 meters long and swung so that it travels in a horizontal, circular path of radius 0.40 m, as shown.
a Calculate the force of tension in the string as the ball swings in a horizontal circle.
b Determine the magnitude of the centripetal acceleration of the ball as it travels in the horizontal circle.
c Calculate the period T (time for one revolution) of the ball’s motion.

Answer 1

Answer:

hey gye mark me as branlist

Answer 2

Answer:

A billiard ball of mass $m=0.15kg$ is attached to a string of length $l=0.50m$ travels through a circular path of radius $r=0.40m$, the force of tension in horizontal direction is $F_{tension}=2.45N$, the magnitude of centripetal acceleration is $a_{c}=13.1m/s^{2}$ and the time period is $T=1.10s$.

Explanation:

We could solve this problem by using a free body diagram and  Newton's law  $F_{net} =ma$

To find force of tension we have to equate net force and centripetal force, since net force acting on billiard ball provide centripetal force.

∑$F=\frac{mv^{2} }{r}$, where $m,v,r$ are mass, velocity, and radius of the circular path respectively.

There are two forces on the ball one is $F_{tension}$ along the string and another one is $F_{g}$ gravitational force due to mass of the ball acting downward, as shown in the figure.

The billiard ball is not accelerating vertically, so we can write along Y direction,

$F_{tension}-F_{g}= 0$

$F_{tension} sin \theta =mg$

Where  $F_{tension} sin \theta$ is the vertical component of tension force.

we can calculate θ using length of the string and circular radius.

$cos \theta =0.40/0.50$

$\theta =\cos ^{-1}\frac{0\cdot 40}{0\cdot 50}=36\cdot 9^{\circ }$

From the question we have, $m=0.150kg, g=9.8 \frac{m}{s^{2} }$

solving for tension force

⇒ $F_{tension} \begin{aligned}\sin 36\cdot 9^0=\left(0\cdot 15\right)^{ }\end{aligned}9.8$

$F_{tension}=2.45N$

The centripetal acceleration can be calculated using horizontal component of tension as it provides centripetal force.

$a_{c}=\frac{F_{tension}cos \theta }{m}$

$a_{c}=\frac{2.45cos36.9}{0.15}=13.1m/s^{2}$

The time for one revolution can be determined by using $T=\frac{distance}{speed}$

$d=2\pi r$ the circumference of the circle

We have centripetal acceleration is $a_{c}=\frac{v^{2} }{r}$

From this speed can be calculated as,$v=\sqrt{ra_{c} }$

Now, the time period is $\frac{2\pi r}{\sqrt{ra_{c} } }$

$T=\frac{2\pi\sqrt{0.40} }{\sqrt{13.1} } =1.10s$