Angular Momentum of a Uniform Rod

[Bob4321]

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1. Homework Statement
A uniform rod of length L1 = 1.5 m and mass M = 2.8 kg is supported by a hinge at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass m is supported by a thin string of length L2 = 1.1 m from the hinge. The particle sticks to the rod on contact. After the collision, θmax= 45°.

a)Find m (in kg)

b) How much energy is dissipated during the collision? (in J)

Homework Equations

I found the question with different numbers but when I plug in mine. It's still wrong.

The Attempt at a Solution

 

[SammyS]

Homework Statement

A uniform rod of length L1 = 1.5 m and mass M = 2.8 kg is supported by a hinge at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass m is supported by a thin string of length L2 = 1.1 m from the hinge. The particle sticks to the rod on contact. After the collision, θmax= 45°.

a)Find m (in kg)

b) How much energy is dissipated during the collision? (in J)

Homework Equations

I found the question with different numbers but when I plug in mine. It's still wrong.
View attachment 85506

View attachment 85501 View attachment 85502 View attachment 85503 View attachment 85504 View attachment 85505

The Attempt at a Solution

Hello Bob4321. Welcome to PF.

Please show your attempt. What did you do in attempting to use that solution?

 

[Bob4321]

Solution Attempt:

Ok so for A:

ω1=((1/3)(2.8)(1.5²))/ ((1/3)(2.8)(1.5²))+m(1.1²)√3(9.81m/s2)/1.5= (4.429kg/s)/(2.1kg)+1.21m

½(4.429kg/s)²/(2.1kg)+1.21m I don't know what to do now. Where did they get the .2 from in their example?

 

[SammyS]

Solution Attempt:

Ok so for A:

ω1=((1/3)(2.8)(1.5²))/ ((1/3)(2.8)(1.5²))+m(1.1²)√3(9.81m/s2)/1.5= (4.429kg/s)/(2.1kg)+1.21m

½(4.429kg/s)²/(2.1kg)+1.21m I don't know what to do now. Where did they get the .2 from in their example?

What is (½)(1 - cos(θ_max) ) ?

It looks like you have an issue regarding "order of operations". The entire expression, (⅓)(2.8)(1.5²)+m(1.1²) should be in the denominator. It's clear that in the solution you're attempting to mimic, the mass, m, is in the denominator.In my opinion: This method of obtaining an answer for an exercise is not to be recommended. If you don't know what quantities are being used, and why they're used, simply getting an answer to work out numerically isn't of much use.

Examining a worked out solution can indeed be very helpful, but don't just try to plug in some numbers into something that you don't understand.

 

[Bob4321]

Ok I have figured out that the .2 is from 1-cos but I still keep getting the wrong answer! Could you solve it so I can see what you did?

 

[Bob4321]

For my problem the (1-cos45) is .48

 

[SammyS]

Ok I have figured out that the .2 is from 1-cos but I still keep getting the wrong answer! Could u solve it so I can see what you did?

That's not how we do things here at PF.

0.48 Is incorrect. Were you using 45 radians rather than 45° ?