# Simple Harmonic Motion of sphere [closed]

Problem: A solid sphere of radius 15cm is pivoted at a point on its surface. It is held so that the diameter passing through both the pivot and the center of the sphere makes a 0.15 radian angle with the vertical. The sphere is released at time = 0.10s.

Find the velocity of the bottom of the sphere at time $t$ = 0.4s.

The equation for the angle between the radius to the pivot and the vertical as a function of time I found:

$\theta_0 = 0.15$

$\theta = 0.15cos(\omega t - \phi)$

$\omega = \sqrt{\frac{g}{7/5R}}$

$\phi = 0.683130511$

Therefore:

$\theta = 0.15cos(6.8313 t - 0.683130511)$

The angular velocity is given by: $\omega = \frac{d^{2}\theta}{dt^{2}} = 1.0247\sin(0.683131 - 6.8313t)$

$\frac{v}{r} = \omega$

$r_b = 0.3cm$

Solving for $v$ I get 0.273m/s. The answer given is 0.14m/s.

## closed as off-topic by John Rennie, Martin, ACuriousMind♦, Kyle Kanos, JimMay 11 '15 at 16:07

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• check value of your angle relation at t=0 – Azad May 11 '15 at 6:36
• what do you mean? The sphere is released at t = 0.10s. So I calculated the phase shift. – Quaxton Hale May 11 '15 at 6:37
• Hi Quaxton and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – John Rennie May 11 '15 at 6:38
• @QuaxtonHale yes my mistake – Azad May 11 '15 at 6:39
• check definition of angular velocity instead – Azad May 11 '15 at 6:40

I've got v(0.4)= 3.066 m/s

I'will use physical pendulum, circular geometry - hyperphysics
with data : radius 0.15 cm , g=9.8 m/s/s The period T for a sphere is 0.09197 sec (last value in the page)
(is more than 10 times faster than a simple pendulum of length 30 cm)
I get $$\theta(t)=\theta_{max}*cos((t-t_{0})*2\pi/T)$$

With the substitution of sin by cos and t by $t-t_0$ we get a maximum 0.15 at time 0.10 s

and at the sagemath I get two graphs $\theta$
|--> 0.150000000000000*cos(10.8724014960424*pi*(2*t - 0.200000000000000)) and $\theta(0.10)$ = 0.150000000000000

and $v(t)=d\theta/dt$
|--> -3.26172044881273*pi*sin(10.8724014960424*pi*(2*t - 0.200000000000000)) and v(0.4) m/s= -3.06576728129694

t,b,g,L,tmax,tstart=var('t,b,g,L,tmax,tstart')
theta = function('theta', t)
theta1 = function('theta1', t)

#b=g/L
#assume(g>0)
#assume(L>0)
L=0.15    ## Length of string
tstart=0.10
g=9.8 ## accelaration due to gravity
tmax=0.15
##### Pendulum's Time Period: ############
#
T=0.091976

theta(t)=tmax*cos((t-tstart)*2*pi/T)
theta
theta(tstart)

#theta1(t)=0.15*cos(6.8313*t-0.683130511)  ## the one of OP
a=plot(theta, 0.10,1.1)
a.show(aspect_ratio=0.5)
#b=plot(theta1, 0.10,1.1)
#b.show(aspect_ratio=0.5)

v=diff(theta,t)
v
print"v(0.4) m/s= ",0.30*v(0.4).n()

b=plot(v, 0.10,1.1)
b.show()