Mechanics (Homework 7)
1.
2.
Part 2: How long does it take $T$ (period) for the ball to rotate once around the axis? Answer in units of seconds.
3.
Part 1. A ball at the end of a string of length 0.81m rotates at a constant speed in a horizontal circle. It makes 7.3 rev/sec.
What is the period of the ball’s motion? Answer in units of sec.
Part 2: What is the frequency of the motion? Answer in units of $Hz$.
Part 3: What is the ball’s angular velocity? Answer in $rad/sec$.
1.
Let angle $\theta$ be the angle formed by the road with the horizontal. In this case by drawing the centrifugal force on the car and the weight of the car one can write
$63 km/h =63000/3600 =17.5 m/s$
$tan(\theta) = F_{cf}/G = (m*v^2)/(R*mg) =v^2/(g*R) =17.5^2/(9.8*55) =0.568$
$\theta = 29.6 degree$
Attention, when drawing the triangle of forces (centrifugal forces, weight) a very common mistake I have met on numerous answers is to write $tan(\theta) = (mg)/F_{cf}$. This is wrong!
2.
1. let L be the length of the thread and $\theta = 24 degree$ the angle made by the rotating thread with the vertical.
the radius of the circle in which the ball moves is $R=L*sin(\theta)$
In the triangle of forces made by the ball weight and the centrifugal force one can write
$tan(\theta) = F_{cf}/G = (m*v^2)/(R*mg) = v^2/(R*g) =v^2/(g*L*sin(\theta))$
$v =\sqrt{g*L*tan(\theta)*sin(\theta)}$
2. The length of the circle in which the ball rotates is $length =2\pi*R =2\pi*L*sin(\theta)$
The period of rotation is
$T = length/V = [2*pi*L*sin(\theta)]/\sqrt{g*L*tan(\theta)*sin(\theta)}$
One must know the length of the thread to compute the numerical values.
The speed is
$v =\sqrt{9.8*3.1*tan(24)*sin(24)} =\sqrt{5.501} = 2.346 m/s$
the period of rotation is
$T = 2*\pi*3.1*sin(24)/2.346 =3.376 s$
3.
1. By definition the period of the movement is the time needed to complete one revolution. Hence it is equal to 1/frequency.
$T = 1/f = 1/7.3 =0,137 sec$
2.the frequency is given directly into the text
$f= 7.3 sec^{-1} =7.3 Hz$
3. The length of the circle is $L =2*\pi*l$ where $l$ is the length of the string. It is completed in $T$ seconds. Hence the speed is simply
$v = L/T = (2*\pi*l)/T =2*\pi*l*f = 2*\pi*0.81*7.3 =37.15 rad/sec$