A pendulum of mass m= 0.8 kg and length l=1 m is hanging from the ceiling. The massless string of the pendulum is attached at point P. The bob of the pendulum is a uniform shell (very thin hollow sphere) of radius r=0.4 m, and the length l of the pendulum is measured from the center of the bob. A spring with spring constant k= 7 N/m is attached to the bob (center). The spring is relaxed when the bob is at its lowest point (θ=0). In this problem, we can use the small-angle approximation sinθ≃θ and cosθ≃1. Note that the direction of the spring force on the pendulum is horizontal to a very good approximation for small angles θ. (See figure)
Take g= 10 m/s2
(a) Calculate the magnitude of the net torque on the pendulum with respect to the point P when θ=5∘. (magnitude; in Nm)
|τP|=
(b) What is the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘? (magnitude; in radians/s2)
|α|=
(c) What is the period of oscillation T of the pendulum? (in seconds)
T=
Is this an 8:01 problem ? If not I will do it.
If it is not I will help.
no it isn't
Thanks
Lets do moments of inertia about the pivot point
First the moment of inertia of this thin shell about a horizontal line through its center is (2/3) mR^2
for derivation see http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html
This has to be transferred up to the pivot point using the parallel axis theorem
I = m L^2 + (2/3) m R^2
here
m = .8
L = 1
R = .4
so
I = .8 + .085 = .885
but how do you get alpha?
Now let's do Moments about the pivot point
let angle theta = T
restoring Force = - k x - m g sin T
but sin t is about T in radians for small T
so
F = -k x - m g T
x is about L T
so
F = -k L T - m g T
moment = M = F L = -(k L^2 + mgL)T
moment = I alpha
-(k L^2 + mgL)T = .885 d^2T/dt^2
well calculate the coefficient of T on the left
-(7 + 8) T = .885 d^2T/dt^2
.885 d^2T/dt^2 = - 15 T
let T = A sin ( w t)
then d^2T/dt^2 = - w^2 T
so
.885 w^2 = 15
w = 2 pi f = 2 pi/period = 4.11
Hey, I am kind of slow and can not do everything at once.
Check arithmetic!
when you do these, use T in RADIANS
5 degrees *pi/180
(a) Calculate the magnitude of the net torque on the pendulum with respect to the point P when θ=5∘. (magnitude; in Nm)
Torque = -15T where T = 5 * pi/180
|τP|=
(b) What is the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘? (magnitude; in radians/s2)
alpha = d^2T/dt^2 = - 15 T /.885
|α|=