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**[SOLVED] Thin rod suspended freely on end and dropped. KE and Center of Mass**

## Homework Statement

A thin rod of length 0.75 m and mass 0.39 kg is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 3.5 rad/s. Neglect friction and air resistance.

(a) Find the rod's kinetic energy at its lowest position

J

(b) Find how far above that position the center of mass rises.

m

## Homework Equations

W=Wo +at

Radian= Radiano + Wo +(1/2)(angular accel)(squared)

W(squared) -Wo(squared) = 2(angular accel)(radians)

Rotational Kinetic Energy equation. K= (1/2)(I)(W)(squared)

W=V(r) where are is radius

Equation for I of thin rod (1/12)(m)(radius)squared

## The Attempt at a Solution

First I tried using the equationf or I of a thin rod and I used the mass of .39 times radius of (.375)squared and then multiplied it by (1/12). Getting I. Then I used the given value of W=3.5rad/s. Using the equation from B2 to find K I got .02799J which is wrong...I have no idea what I did wrong here.

Part B: Working the center of mass portion of this problem I assumed the mass was even distributed and would be toward the center of the pole. If the pole is vertical, as it is at the bottom of its spin then the center of mass should be halfway up the pole. So I got .375m or half the radius for this part. I got this wrong also. Any help would be greatly appreciated.