Mechanics (Homework 7)


A curve or radius 55 m is banked so that a car traveling w/uniform speed 63 km/hr can round the curve w/out relying on friction to keep from sliding to the left or right.  Acceleration of gravity is 9.8m/s2.  What is the angle $\theta$? Answer in degrees.


Two Part Question: Part 1: A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9.8 m/s2. What is the ball’s speed when it is in circular motion? Answer in units of m/s.

Part 2: How long does it take $T$ (period) for the ball to rotate once around the axis? Answer in units of seconds.


Three Part Question:

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 $rad/sec$.


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!


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

$T = 2*\pi*3.1*sin(24)/2.346 =3.376 s$


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

$v = L/T = (2*\pi*l)/T =2*\pi*l*f  = 2*\pi*0.81*7.3 =37.15 rad/sec$