- #1

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## Homework Statement

## Homework Equations

Torque equations and Force equations

Moment of inertia for disc rotating around centre: (mr^2)/2

After using parallel axis theorem, the I about the edge is I = (3/2)mr^2

Thank you,

Aralox

- Thread starter Aralox
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- #1

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Torque equations and Force equations

Moment of inertia for disc rotating around centre: (mr^2)/2

After using parallel axis theorem, the I about the edge is I = (3/2)mr^2

Thank you,

Aralox

- #2

- 12,115

- 151

Using the parallel axis theorem is a good start. Can you do the following:## Homework Equations

Torque equations and Force equations

Moment of inertia for disc rotating around centre: (mr^2)/2

After using parallel axis theorem, the I about the edge is I = (3/2)mr^2

Thank you,

Aralox

1. Given I, and the torque about the pivot point, can you calculate the angular acceleration of the disk?

2. From the angular acceleration about the pivot, you should be able to figure out the acceleration of the disk's center of mass.

- #3

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I used Net Torque = mgr, as initially the weight is perp. to radius to pivot.

Since net torque = Iα (α is angular acc.),

mgr = Iα.

Since I = (3/2) mr^2, I simplified the previous equation to

α = 2g/3r, which happens to be a constant, so it looks like i have the initial angular acceleration of the disc about the pivot at the edge.

To find this, i used the eqation a = αr,

so acceleration of disc's center: a = (2g/3r) * r = (2/3)g

How would i find the vertical reaction of the pin from this though?

Just brainstorming, could i do something where i took the disc to be not moving and the world rotating around it, meaning i would have the edge moving at a = (2/3)g, which i could get some kind of reaction force from?

This interpretation/application step is what im having the most trouble with.

Thanks,

Aralox

PS: Just a test to figure out this 'latex reference' thing

[itex]\sum Torque[/itex]

- #4

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Well, now that you know the acceleration of the disk's center of mass, you pretty much know what the.

.

.

...so acceleration of disc's center: a = (2g/3r) * r = (2/3)g

How would i find the vertical reaction of the pin from this though?

Just brainstorming, could i do something where i took the disc to be not moving and the world rotating around it, meaning i would have the edge moving at a = (2/3)g, which i could get some kind of reaction force from?

This interpretation/application step is what im having the most trouble with.

- #5

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[itex]\sum F[/itex] = N - mg = ma,

which leads to N = (5/3)mg

would N be the 'vertical component of reaction at the pin'? Isnt N the force (reaction force of some sort) vector originating at the centre of the disc, since im taking net force at the centre?

- #6

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Just a little bump :)

- #7

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Well,Alrighty, so if i had:

[itex]\sum F[/itex] = N - mg = ma,

which leads to N = (5/3)mg

Yes.would N be the 'vertical component of reaction at the pin'?

No,Isnt N the force (reaction force of some sort) vector originating at the centre of the disc, since im taking net force at the centre?

- #8

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Thank you for all the help Redbelly98! :)

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