A pendulum of mass m= 0.9 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= 12 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|=
unanswered
(b) What is the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘? (magnitude; in radians/s2)
|α|=
unanswered
(c) What is the period of oscillation T of the pendulum? (in seconds)
T=
8 answers
8:01 final question. The easiest one.
damon can u please tell me about the T that is the last part of this question
omega= sqrt((kl^2+mgl)/I)
please tell me the exact expression for this question.
|ôP|= 1.83
|á|= 1.84
T= 1.37
|á|= 1.84
T= 1.37
Can you plz give out the formula??
I = m L^2 + (2/3) m R^2
-(k L^2 + mgL)T = .885 d^2T/dt^2
well calculate the coefficient of T on the left e.g.
-(7 + 8) T = .885 d^2T/dt^2
.885 d^2T/dt^2 = - 15 T
Torque = -15T where T = 5 * pi/180
alpha = d^2T/dt^2 = - 15 T /.885
-(k L^2 + mgL)T = .885 d^2T/dt^2
well calculate the coefficient of T on the left e.g.
-(7 + 8) T = .885 d^2T/dt^2
.885 d^2T/dt^2 = - 15 T
Torque = -15T where T = 5 * pi/180
alpha = d^2T/dt^2 = - 15 T /.885
.885 w^2 = 15
w = 2 pi f = 2 pi/period = 4.11 for example and you solve fot period
w = 2 pi f = 2 pi/period = 4.11 for example and you solve fot period