Finding moment of inertia of a pulley

[bologna121121]

Homework Statement

Two masses, one of $$m_{1}= 3kg$$, the other of $$m_{2}=4kg$$, hang from opposite sides of a pulley of radius .15m. When released from rest, the heavier mass falls .34m in 4s. What is the rotational inertia of the pulley?

Homework Equations

Newton's second law and $$\tau = I\alpha$$

The Attempt at a Solution

I tried using Newton's 2nd law on each of the masses:
$$T_{1} - m_{1}g = m_{1}a$$
$$m_{2}g - T_{2} = m_{1}a$$
where $$T_{1}$$ and $$T_{2}$$ are the tensions of the rope on each mass.
solving for these tensions yields
$$T_{1} = m_{1}(g+a)$$
$$T_{2} = m_{2}(g-a)$$
I originally planned to plug these into $$\tau = I\alpha$$ and use linear kinematics equations to solve for the acceleration and therefore the angular acceleration, but it is clear that $$T_{1}$$ is greater than $$T_{2}$$ something seems to have gone wrong, because wouldn't this imply that, through Newton's third law, the net torque would be causing $$m_{1}$$, the lighter block, to fall? Thanks for the help.

 

[RK7]

My approach would be:
Find the acceleration of the block and relate this to the angular acceleration of the pulley then find the resultant torque on the pulley. You don't need to worry about tensions.

 

[bologna121121]

But don't the tensions determine the resultant torque on the pulley?

 

[RK7]

Edit: Yeah sorry I'm talking rubbish, I'll have another think tomorrow

 

[gneill]

Homework Statement

Two masses, one of $m_{1}= 3kg$, the other of $m_{2}=4kg$, hang from opposite sides of a pulley of radius .15m. When released from rest, the heavier mass falls .34m in 4s. What is the rotational inertia of the pulley?

Homework Equations

Newton's second law and $\tau = I\alpha$

The Attempt at a Solution

I tried using Newton's 2nd law on each of the masses:
$T_{1} - m_{1}g = m_{1}a$
$m_{2}g - T_{2} = m_{1}a$
where $T_{1}$ and $T_{2}$ are the tensions of the rope on each mass.
solving for these tensions yields
$T_{1} = m_{1}(g+a)$
$T_{2} = m_{2}(g-a)$
I originally planned to plug these into $\tau = I\alpha$ and use linear kinematics equations to solve for the acceleration and therefore the angular acceleration, but it is clear that $T_{1}$ is greater than $T_{2}$ something seems to have gone wrong, because wouldn't this imply that, through Newton's third law, the net torque would be causing $m_{1}$, the lighter block, to fall? Thanks for the help.

Actually, you're doing okay. T1 will be less that T2 because the mass multipliers make a difference.

You can determine the acceleration form the given information: "the heavier mass falls .34m in 4s".

Once you have the acceleration you can determine the tensions T1 and T2. You'll only need the equation for the motion of the pulley to complete.

 

[Genoseeker]

Homework Statement

Two masses, one of $$m_{1}= 3kg$$, the other of $$m_{2}=4kg$$, hang from opposite sides of a pulley of radius .15m. When released from rest, the heavier mass falls .34m in 4s. What is the rotational inertia of the pulley?

Homework Equations

Newton's second law and $$\tau = I\alpha$$

The Attempt at a Solution

I tried using Newton's 2nd law on each of the masses:
$$T_{1} - m_{1}g = m_{1}a$$
$$m_{2}g - T_{2} = m_{1}a$$
where $$T_{1}$$ and $$T_{2}$$ are the tensions of the rope on each mass.
solving for these tensions yields
$$T_{1} = m_{1}(g+a)$$
$$T_{2} = m_{2}(g-a)$$
I originally planned to plug these into $$\tau = I\alpha$$ and use linear kinematics equations to solve for the acceleration and therefore the angular acceleration, but it is clear that $$T_{1}$$ is greater than $$T_{2}$$ something seems to have gone wrong, because wouldn't this imply that, through Newton's third law, the net torque would be causing $$m_{1}$$, the lighter block, to fall? Thanks for the help.

you are forgotting what T1 and T2 are actually doing..

T1 > T2

that part is clear. but...

T1 is increase net acc of the system by pulling m1 upwards.
T2 is decrease net acc of the system...aka...slowing down m2...by pulling it upwards.

So when T1 > T2 that means the system rotates in the direction of m2. the larger mass.

I hope that clears it up for you.

 

[Genoseeker]

Actually, you're doing okay. T1 will be less that T2 because the mass multipliers make a difference.

You can determine the acceleration form the given information: "the heavier mass falls .34m in 4s".

Once you have the acceleration you can determine the tensions T1 and T2. You'll only need the equation for the motion of the pulley to complete.

that is kinda wrong.

T1>T2

think of it like this.

T2 is a force caused by m1g on m2
T1 is a force caused by m2g on m1

 

[gneill]

that is kinda wrong.

T1>T2

think of it like this.

T2 is a force caused by m1g on m2
T1 is a force caused by m2g on m1

If T1 were greater than T2, then the pulley would turn towards the T1 side and m2 would rise, not fall.

 

[azizlwl]

$$T_{1} = m_{1}(g+a)$$
$$T_{2} = m_{2}(g-a)$$
I originally planned to plug these into $$\tau = I\alpha$$ and use linear kinematics equations to solve for the acceleration and therefore the angular acceleration, but it is clear that $$T_{1}$$ is greater than $$T_{2}$$ something seems to have gone wrong, because wouldn't this imply that, through Newton's third law, the net torque would be causing $$m_{1}$$, the lighter block, to fall? Thanks for the help.

T1 is not greater than T2.

T2-T1= m2g -m2a -m1g-m1a=g(m2-m1)-a(m2-m1)>0

 

[azizlwl]

$$T_{1} = m_{1}(g+a)$$
$$T_{2} = m_{2}(g-a)$$
I originally planned to plug these into $$\tau = I\alpha$$ and use linear kinematics equations to solve for the acceleration and therefore the angular acceleration, but it is clear that $$T_{1}$$ is greater than $$T_{2}$$ something seems to have gone wrong, because wouldn't this imply that, through Newton's third law, the net torque would be causing $$m_{1}$$, the lighter block, to fall? Thanks for the help.

T1 is no greater than T2.

T2-T1= m2g -m2a -m1g-m1a=g(m2-m1)-a(m2+m1)>0