Rotation with a Pulley and Masses

[BlasterV]

one block has a mass M = 500 g, the other has mass m = 460 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.00 cm. When released from rest, the heavier block falls 85.5 cm in 2.47 s (without the cord slipping on the pulley).

First I did conversions:
M = .5kg, m = .46 kg, r = .05 m
fall = .855m in 2.47 s

I am unable to do part b and c right, so I am unable to go further.
Here is my work:

.855m = 1/2 a (2.47^2) a = .28 m/s^2

however, It asks for 2 tensions. I never learned how to separate them like that, especially with a pulley. I did:

9.8 m/s^2 * .5kg = 4.9 N
9.8 m/s^2 * .46kg = 4.508 N

Tension of Cord = 4.9N + 4.508N. Obviously this isn't right, where do I go from here, or did I miss something?


(a) What is the magnitude of the block's acceleration?
.28 m/s2
(b) What is the tension in the part of the cord that supports the heavier block?
N
(c) What is the tension in the part of the cord that supports the lighter block?
N
(d) What is the magnitude of the pulley's angular acceleration?
rad/s2
(e) What is its rotational inertia?
kg · m2

 

[BlasterV]

I got it all, thanks though

 

[thepatientmental]

Your calculations for the acceleration of the block and the tensions in the cord are correct. However, to find the tensions in the cord that support each block separately, you need to use the concept of torque.

Torque is the rotational equivalent of force, and it is calculated by multiplying the force by the distance from the pivot point. In this case, the pivot point is the center of the pulley.

To find the tension in the part of the cord that supports the heavier block, we can use the equation:

Tension = (mass of the block) * (acceleration of the block) + (mass of the pulley) * (angular acceleration of the pulley) * (radius of the pulley)

T = (.5 kg)(.28 m/s^2) + (.5 kg)(a)(.05 m)
T = .14 N + .025a N

Similarly, to find the tension in the part of the cord that supports the lighter block, we can use the equation:

Tension = (mass of the block) * (acceleration of the block) - (mass of the pulley) * (angular acceleration of the pulley) * (radius of the pulley)

T = (.46 kg)(.28 m/s^2) - (.5 kg)(a)(.05 m)
T = .129 N - .025a N

To find the angular acceleration of the pulley, we can use the equation:

Torque = (moment of inertia) * (angular acceleration)

Since the pulley is a uniform disk, its moment of inertia can be calculated as (1/2) * (mass of the pulley) * (radius of the pulley)^2. Therefore:

(.14 N + .025a N)(.05 m) = (.5 kg)(1/2)(.05 m)^2 * a
.007 Nm = .0125 kgm^2 * a
a = .56 rad/s^2

To find the moment of inertia of the pulley, we can use the equation:

Moment of inertia = (mass of the pulley) * (radius of the pulley)^2

I = (.5 kg)(.05 m)^2
I = .00125 kgm^2

So, the magnitude of the pulley's angular acceleration is .56 rad/s^2 and its rotational inertia is .00125 kgm^2.