Thin rod suspended freely on end and dropped. KE and Center of Mass

[rosstheboss23]

[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.

 

[hage567]

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

For (a), you are using the wrong formula for I. The one you have is for a rod rotating about its centre. You need the one were it is rotating about one end. They are not the same.

 

[rosstheboss23]

Ok thanks. I will look up the formula, but I am not sure if its in my textbook. Does everything else look reasonably good? Also I was wondering if you knew anything on part B. I really appreciate your help.

 

[hage567]

Everything looks OK except for your I value.

For part (b): You are trying to find the height that the rod comes to rest after it swings through the lowest point. Apply conservation of energy.

 

[rosstheboss23]

Ok I got part A. Formula for I. For those interested is (1/3)(m)(L(squared)). This is for a rod rotating about one end. Part B is I am still stuck on. Any advice?

 

[rosstheboss23]

Ok. I will do so but for the potential energy part of conservation I would still use mgh right?

 

[hage567]

Yes, that's right.

 

[rosstheboss23]

Ok. Would the conservation of energy for angular momentum go like this mgh =.5(1/3)(m)(L(squared))(3.5)(squared) + mgh(final). I tried this and got (.39)(9.8)(.75)=.4479(Part a answer) +(.39)(9.8)(height above bottom pt.) and I got .6328m. I find the concept hard to understand. Is this thin rod being held horizontal and then one side is dropped until it is completely vertical?

 

[hage567]

I find the concept hard to understand. Is this thin rod being held horizontal and then one side is dropped until it is completely vertical?

Well, one end is being released but you can't say if its horizontal at that time. It doesn't really matter for this question. You are trying to find the height that the centre of the rod swings up to. Taking the centre of mass (so the middle of the rod in this case) as your reference point when the rod goes through the vertical, you can say the G.P.E. there is zero; the rod only has rotational kinetic energy. So that energy must all be converted into G.P.E. if the rod momentarily stops at its highest point.

You are on the right track I think, you just need to get a clear picture of what is being asked in the question.

 

[rosstheboss23]

Ok. Thanks I just found out the idea with this one. KE=mgh so .4479J/[(.39kg)(9.8m/s(squared)]. So the answer would be .1172m. Thanks again for all your help.

 

[hage567]

You're welcome.