Pulley and torque homework

  • #1

Homework Statement


The two blocks in the figure (Intro 1 figure) are connected by a massless rope that passes over a pulley. The pulley is 17 cm in diameter and has a mass of 2.0 kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.46 Nm .

If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

(Intro 1 Figure):http://session.masteringphysics.com/problemAsset/1070551/3/12.P70.jpg


Homework Equations



torque= r * F
sum of torques = moment of inertia * angular acceleartion
angular acceleration * radius = acceleration
x=v0t + 1/2 a t^2

The Attempt at a Solution



I now that v0 is o so the i solve for t in the equation, t = sqrt(2h / a).
a = [ F / r + (m1 - m2)g ] / (m1 + m2 + m3 / 2)

then i plug in a into t=sqrt(2h/a)
= sqrt{ h(2m1 + 2m2 + m3) / [ F / r + (m1 - m2)g ] }
=t = sqrt{ 1( 2 * 4 + 2 * 2 + 2) / [ 0.46 / 0.17 + (4 - 2)9.81 ] }
=.79s
I got it wrong, can anyone help me to see what i did wrong.
 

Answers and Replies

  • #2
11
0
The diameter is 17cm, so your radius is half of that. Give that a try. I see you have 17 instead for radius.
 
  • #3
Doc Al
Mentor
44,990
1,263
a = [ F / r + (m1 - m2)g ] / (m1 + m2 + m3 / 2)
Show how you derived this result.
 
  • #4
This is how i found the equation for acceleration

Let:
m1 be the left hand mass,
T1 be the left hand tension,
m2 be the right hand mass,
T2 be the right hand tension,
g be the acceleration due to gravity,
a be the acceleration of the masses,
alpha be the angular acceleration of the pulley,
r be the radius of the pulley (assumed solid),
F be the friction couple,
h be the descent for m1,
t be the time taken.

For mass m1:
m1 g - T1 = m1 a
T1 = m1(g - a)

For mass m2:
T2 - m2 g = m2 a
T2 = m2(g + a)

For the pulley:
(T1 - T2)r - F = I alpha
= (m3 r^2 / 2)(a / r)

T1 - T2 = F / r + m3 a / 2

Eliminating T1 and T2:
m1(g - a) - m2(g + a) = F / r + m3 a / 2

(m1 - m2)g - (m1 + m2)a - m3 a / 2 - F / r
(m1 + m2 + m3 / 2)a = F / r + (m1 - m2)g

a = [ F / r + (m1 - m2)g ] / (m1 + m2 + m3 / 2)

Did i derieved it correctly?


Also did i solve for time corectly.
h = at^2 / 2
t = sqrt(2h / a)
= sqrt{ h(2m1 + 2m2 + m3) / [ F / r + (m1 - m2)g ]


Ok i see that i plug in the wrong number for the radius.
=t = sqrt{ 1( 2 * 4 + 2 * 2 + 2) / [ 0.46 / 0.085 + (4 - 2)9.81 ] }
t=.7478 s

is this answer correct?
 
  • #5
Doc Al
Mentor
44,990
1,263
Eliminating T1 and T2:
m1(g - a) - m2(g + a) = F / r + m3 a / 2
OK.
(m1 - m2)g - (m1 + m2)a - m3 a / 2 - F / r
(m1 + m2 + m3 / 2)a = F / r + (m1 - m2)g
Redo this last step.
 
  • #6
I dont understand wat you mean, did i made a mistake deriving it? i did over and i still got the same formula.
 
  • #7
343
1
hmm I got

2g-Ff=ma and Ff*r=I*(a/r), where: a and Ff is unknown. how is this wrong?

you can't look at it like one object anymore?
 
Last edited:
  • #8
Doc Al
Mentor
44,990
1,263
I dont understand wat you mean, did i made a mistake deriving it? i did over and i still got the same formula.
Everything you did up to and including where I said "OK" was good. You just made an algebraic mistake where I said "Redo this last step".
 
  • #9
Thanks for your help, i figure out where i made an algebra mistake
 

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