let x = C-r
height of center of ball = x(1 - cos Theta)
for small Theta
1 - cos Theta = 1-1+theta^2/2 =theta^2/2
Potential energy = m g x(theta^2)/2
ke of ball about top point = (1/2)mx^2 (d theta/dt)^2
ke of ball around its center
= (1/2) I omega^2
I =(2/5) m r^2
omega= (x/r)d theta/dt
ke = (1/2)(2/5)mr^2 (x^2/r^2) (d theta/dt)^2
theta = A sin w t
d theta/dt = A w cos wt
max Potential = max kinetic
m g x(A^2)/2= A^2w^2 (1/5)m x^2
w^2 = g(5/2x)
w = 2 pi f = 2 pi/T
w^2 =(2 pi)^2/T^2
T^2 = (2 pi)^2 /w^2
= (2 pi)^2 [2(C-r)]/5g
A ball rolls forwards and backwards on a curved track as shown in Fig. 1.1.
ball
flexible track
Fig. 1.1
It is suggested that the period T of the oscillations is related to the radius r of the ball and the
radius of curvature C of the track by the relationship
T 2 = 28π2
5g (C – r )
where g is the acceleration of free fall.
You are provided with a flexible track. Design a laboratory experiment to test the relationship
between T and r. Explain how your results could be used to determine a value for C. You
should draw a diagram, on page 3, showing the arrangement of your equipment. In your
account you should pay particular attention to
(a) the procedure to be followed,
(b) the measurements to be taken,
(c) the control of variables,
(d) the analysis of the data,
(e) the safety precautions to be taken.
2 answers
Left out ke about pin at top
let x = C-r
height of center of ball = x(1 - cos Theta)
for small Theta
1 - cos Theta = 1-1+theta^2/2 =theta^2/2
Potential energy = m g x(theta^2)/2
ke of ball about top point = (1/2)mx^2 (d theta/dt)^2
ke of ball around its center
= (1/2) I omega^2
I =(2/5) m r^2
omega= (x/r)d theta/dt
ke = (1/2)(2/5)mr^2 (x^2/r^2) (d theta/dt)^2
theta = A sin w t
d theta/dt = A w cos wt
max Potential = max kinetic
m g x(A^2)/2= A^2w^2 (1/5)m x^2+(1/2)m x^2 A^2 w^2
g /2x = w^2 (1/5+1/2) = w^2(7/10)
w^2 = g(5/7x)
w = 2 pi f = 2 pi/T
w^2 =(2 pi)^2/T^2
T^2 = (2 pi)^2 /w^2
= (2 pi)^2 [7(C-r)]/5g
= 28 pi^2 (C-r)/5g
let x = C-r
height of center of ball = x(1 - cos Theta)
for small Theta
1 - cos Theta = 1-1+theta^2/2 =theta^2/2
Potential energy = m g x(theta^2)/2
ke of ball about top point = (1/2)mx^2 (d theta/dt)^2
ke of ball around its center
= (1/2) I omega^2
I =(2/5) m r^2
omega= (x/r)d theta/dt
ke = (1/2)(2/5)mr^2 (x^2/r^2) (d theta/dt)^2
theta = A sin w t
d theta/dt = A w cos wt
max Potential = max kinetic
m g x(A^2)/2= A^2w^2 (1/5)m x^2+(1/2)m x^2 A^2 w^2
g /2x = w^2 (1/5+1/2) = w^2(7/10)
w^2 = g(5/7x)
w = 2 pi f = 2 pi/T
w^2 =(2 pi)^2/T^2
T^2 = (2 pi)^2 /w^2
= (2 pi)^2 [7(C-r)]/5g
= 28 pi^2 (C-r)/5g