Question

The acceleration of the system is...

Answer 1

A. 1/3 g
B. 4/13 g
C. 1/10 g
D. 2/13 g
E. 11/13 g

Answer 2

Ok, this is similar to an earlier problem. The first thing that needs to be done is to draw a force diagram. What are the forces? There's a gravitational force acting on m1, a gravitational force acting on m2, AND there's a force that's turning the pulley. Let's look at that first. Torque is given by:\[\tau =I \alpha =Fr\]but want we want is a force, so let's divide both sides by r:\[\frac{ I \alpha }{ r }=F _{torque}\]Now angular acceleration is:\[\alpha = \frac{ a _{tangential} }{ r }\]where α is the angular acceleration; and the a is the tangential acceleration, i.e. the acceleration caused by the rope, i.e. the net acceleration of the system. So with that in hand, the overall force resulting from the torque is:\[F _{torque}=\frac{ Ia _{tangential} }{ r ^{2} }=\frac{ Ia _{t} }{ r ^{2} }\]Great. Now back to those summed forces, what you should have should look like this:\[F _{net}=F _{m _{2}}-F _{m _{1}}-F _{torque}\]It should make sense from looking at the problem that the force due to 1 works in the opposite direction of m2. Likewise the same is true of the force due to the torque on the pulley. If we now get explicit, the resulting formula is:\[m _{total}a _{net}=m _{2}g-m _{1}g-\frac{ Ia _{net} }{ r ^{2} }\]Note that the tangential acceleration of the pulley is equal to the net acceleration of the system.

Answer 3

Does that make sense?

Answer 4

Note that all of the answers are in terms of g, so that should tell you to leave g in your equations as g. Don't substitute in its numerical value.

Answer 5

m2 is 6 kg
m1 is 4 kg
Ialpha is 12 kg
r = 2

\[6-2-\frac{ 12 }{ 2^2 }\]

Am I heading the right direction?

Answer 6

Things don't look quite right. Let's solve the equation I gave you for a_net:\[a _{net}=\frac{ g(m _{2}-m _{1}) }{ (m _{total}+\frac{ I }{ r ^{2} }) }\]You should try to solve the overall equation to see if you get the same solution for a_net.

Answer 7

\[\frac{ 9(6-4) }{10+\frac{ 12 }{ 2^2 } }\] Like this?

Answer 8

That looks good, except remember that all of the answers are in terms of g. They are not purely numerical answers.

Answer 9

So should I convert to g before or after solving the problem?

Answer 10

You don't need to do any conversion. Just change the 9 to g.

Answer 11

Ahh ok thanks again