- #1

Adriano25

- 40

- 4

## Homework Statement

In the drawing below, a rod of length L and mass M is pivoted a distance L/4 from one end. The pivot attaches the rod to a smooth horizontal table, allowing the rod to rotate frictionlessly in a horizontal plane (so that gravity does not affect the motion). The end furthest from the pivot is attached to an unstretched spring, and the other end of the spring is attached to a wall. The rod is rotated counterclockwise through 0.30 rad and released at time t = 0.

(a) Starting from a fundamental equation for rotational motion, derive an equation which shows that the oscillation of the rod is an example of simple harmonic motion.

(b) By inspection of the equation you derived in part (a), write down an equation for the angular frequency of the rotation.

(c) If M = 0.70 kg, k = 20.0 N/m, and L = 0.40 m, write down an equation which gives the angle between the rod and its original orientation as a function of time. Give numerical values for any parameters appearing in your expression.

(d) What is the angular velocity of the rod 0.25 s after its release?

## Homework Equations

I worked through part a) and b) and I got that the angular frequency is equal to:

ω=sqrt(27k/7m)

k=spring constant

m=mass of the rod

This answer agrees with the answer sheet. I'm just having troubles for part c) & d)

## The Attempt at a Solution

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ω=sqrt[(27)(20N/m)/(7)(0.7kg)]

ω = 10.5 s

^{-1}

For part c)

I found the equation for position as a function of time to be:

θ(t) = Acos(ωt)

In which we would need to substitute ω for the values we found in part b)

For part d)

The angular velocity as a function of time is:

ω(t) = -Aωsin(ωt)

Thus, plugging in all the values should give me the angular velocity at 0.25 sec.

ω(t) = -(0.3rad)(10.5s

^{-1})sin(10.5s

^{-1}*(0.25s)

ω(t) = -1.56 rad/s

The answer sheet provides an answer of -0.735 rad/s.

Am I doing something wrong or is this result in the answer sheet incorrect?

Thank you