Linear and angular acceleration problems.

  • Thread starter brenfox
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  • #1
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Homework Statement



A mass of 0.5kg is suspended from a flywheel. If the mass is released from rest and falls a distance of 0.5m in 1.5s. ( mass of wheel = 3kg, Radius = 0.3m, Radius of gyration = 212mm ).

Calculate : a:The linear acceleration of the mass?
b: The angular acceleration of the wheel?


Homework Equations



v=s/t
a= v-u/t
s=ut+1/2at^2

The Attempt at a Solution



I am getting conflicting answers to these questions. Heres my attempt!
v = s/t so... v = 0.5/1.5 = 0.3333ms-1.
Then...
a = v-u/t so... v = 0.3333 - 0/1.5 = 0.2222ms-2

So linear acceleration = 0.2222ms-2

Question b : angular acceleration = acceleration/radius.

so...

angular acceleration = 0.2222/ 0.3 = 0.74 Rads-2.

Now my linear acceleration is 0.222ms-2

so if i put this answer into the equation s = ut + 1/2at^2. s should equal 0.5m. so..

s = 1.5+1/2X0.2222x1.5^2 = 1.75s??? The answer should be 0.5m?

Any help will be greatfully accepted.
 

Answers and Replies

  • #2
33
3
In your first equation you have not considered that there is acceleration so v is not s/t

Edit: Also you have not considered that it begins from REST and falls 0.5m in 1.5s
 
  • #3
71
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Average velocity = 0.5/1.5 = 0.3333ms

So the distance from rest to fall = 0.5m so...

0.333/0.5 = 0.6666ms.

So velocity increases from 0 to 0.666ms in 1.5s.

then: a= 0.6666/1.5 = 0.444ms-2 is the linear acceleration????
 
  • #4
33
3
No. You have acceleration so
v = at
s = 0.5at^2 are your equations
 
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  • #5
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third time lucky

v = at

a = v/t

a = 0.3333/1.5 = 0.2222ms-1

a = 0.2222ms-1 but distance is 0.5m so:

0.2222/0.5 = 0.4444ms-2

linear acceleration = 0.4444ms-2.

checking with equation s = 0.5at^2

s = 0.5x0.444x1.5^2 = 0.5m

so inserting 0.4444 into the above equation gives me the correct distance. To my knowledge this tells me 0.4444ms-2 is correct??
 
  • #6
CAF123
Gold Member
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third time lucky

v = at

a = v/t

a = 0.3333/1.5 = 0.2222ms-1

a = 0.2222ms-1 but distance is 0.5m so:

0.2222/0.5 = 0.4444ms-2

linear acceleration = 0.4444ms-2.

checking with equation s = 0.5at^2

s = 0.5x0.444x1.5^2 = 0.5m

so inserting 0.4444 into the above equation gives me the correct distance. To my knowledge this tells me 0.4444ms-2 is correct??
It is correct now. In your previous post, your 'a' had dimensions of 1/t2, so the answer before was coincidental.
 
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  • #7
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Well done. I think you are on your start doing physics. Try making good shapes and for every body you study draw the forces upon it so you can see how it will react. ;)
 
  • #8
71
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Angular acceleration = acceleration/radius

so

0.444/0.3 = 1.48 rads-2

This looks too simple to be correct!
 
  • #9
33
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This is valid. Angular and linear velocities and accelerations are related simply with the radius
 
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