Asked by Robert C

Suppose that a simple pendulum consists of a small 66.0 g bob at the end of a cord of negligible mass. Suppose that the angle between the cord and the vertical is given by
θ = (0.0800 rad) cos[(4.30 rad/s)t + ϕ]

(a) What is the pendulum's length?

(b) What is its maximum kinetic energy?
Answered by Robert C
I figured out a) by using w=2pi/T
T=1.46 then used T to solved T=(2pi*sq root L/g) L = .53 but I can't figure out an equation to give me the velocity. Any ideas?
Answered by drwls
(a) w = 4.40 rad/s
is the angular frequency, sqrt(g/L)

Solve for L, the pendulum length

L = g/w^2

(b) Max velocity Vmax = = L*theta*w

Maximum KE = (1/2) M Vmax^2
Answered by drwls
You are right about L = 0.53. Don't forget the units (meters)

In simple harmonic motion,
max velocity = w*(Amplitude) and
max acceleration = w^2*(Ampitude)

In your case I had to add an L factor to get linear velocities from the angular amplitude.
Answered by Robert C
Where did the Vmax = L*theta*w
come from? How do I find theta?
Answered by drwls
Theta should be theta-max, the angular amplitude, which is 0.0800 radians

Sorry about that
Answered by Robert C
Vmax= (.53m)(.0800rad)(4.30 rad/s)
=.182

KE=1/2mv^2
=1/2(.066kg)(.182)^2
=.0011 or 1.09e-03
That doesn't look right to me. what am I doing wrong? ;_;
Answered by Robert C
Another question, sorry, trying my bests to understand your thinking. I don't see where you knew to multiple L * theta * W.
I understand that the 2nd derivative of the equation gives you the Velocity function.
Answered by drwls
No, the first derivative of your theta vs t (multiplied by L) gives you the velocity function.

L*theta_max is the displacement amplitude (in small angle approximation) ; so
w*L*theta_max
is the maximum velocity


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