MIT OCW 8.01 PS10.6: A Massive Pulley and a Block on an Incline

  • #1
giodude
30
1
Homework Statement
(Screen shot of question is posted below)

Consider a pulley of mass ##m_{p}## and radius ##R## that has a moment of inertia ##\frac{1}{2}m_{p}R^{2}##. The pulley is free to rotate about a frictionless pivot at its center. A massless string is wound around the pulley and the other end of the rope is attached to a block of mass ##m## that is initially held at rest on a frictionless inclined plane that is inclined at an angle ##\beta## with respect to the horizontal. The downward acceleration of gravity is ##g##. The block is released from rest.

How long does it take the block to move a distance ##d## down the inclined plane? Write your answer using some or all of the following: ##R##, ##m##, ##g##, ##d##, ##m_{p}##, ##\beta##.
Relevant Equations
$$\tau_{total} = I_{s} \alpha$$
$$I_{s} = \frac{1}{2}m_{p}R^{2}$$
$$a_{1} = \alpha_{1}R$$
Set up the force equations:
(1) ##mgsin(\beta) - T = ma_{1}##
(2) ##TR = I_{s}\alpha_{1}##

Multiply (1) by ##R## and isolate ##TR##:
##R(mgsin(\beta) - T) = R(ma_{1})##
##mRgsin(\beta) - TR = mRa_{1}##
##TR = mRgsin(\beta) - mRa_{1}##

Plug ##TR## into (2):
##TR = I_{s}\alpha_{1}##
##mRgsin(\beta) - mRa_{1} = I_{s}\alpha_{1}##
##mRgsin(\beta) - mRa_{1} = \frac{1}{2}m_{p}R^{2}\alpha_{1}##

Solve for the ##\alpha_{1}##:
##\alpha_{1} = \frac{mRgsin(\beta) - mRa_{1}}{\frac{1}{2}m_{p}R^{2}}##
(3) ##\alpha_{1} = 2\frac{mgsin(\beta) - ma_{1}}{m_{p}R} = \frac{a_{1}}{R}##

We now use (3) to solve for linear acceleration, ##a_{1}##:
##2\frac{mgsin(\beta) - ma_{1}}{m_{p}R} = \frac{a_{1}}{R}##
##m_{p}a_{1} + 2ma_{1} = 2mgsin(\beta)##
(4) ##a_{1} = \frac{2mgsin(\beta)}{m_{p} + 2m}##

Use (4) to solve the linear kinematics equation for t:
##d = \frac{1}{2} a_{1} t^{2}##
##t = \sqrt{\frac{2d}{a_{1}}}##
$$t = \sqrt{\frac{(m_{p} + 2m)d}{mgsin(\beta)}}$$

I wonder if this solution is correct given that the time to move distance ##d## is independent of the radius ##R## of the pulley. The only intuitive possibility I could think of is a proportional relationship between torque required to achieve a certain angular acceleration and the size of the pulley. Since as the pulley increases as would the leverage and torque required to achieve same angular acceleration. However, I'm not confident in this intuition so I'd love feedback on (a) if my solution is correct and (b) if my intuition explaining the solution is correct. Thank you in advance!
 

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  • #2
giodude said:
Homework Statement: (Screen shot of question is posted below)

Consider a pulley of mass ##m_{p}## and radius ##R## that has a moment of inertia ##\frac{1}{2}m_{p}R^{2}##. The pulley is free to rotate about a frictionless pivot at its center. A massless string is wound around the pulley and the other end of the rope is attached to a block of mass ##m## that is initially held at rest on a frictionless inclined plane that is inclined at an angle ##\beta## with respect to the horizontal. The downward acceleration of gravity is ##g##. The block is released from rest.

How long does it take the block to move a distance ##d## down the inclined plane? Write your answer using some or all of the following: ##R##, ##m##, ##g##, ##d##, ##m_{p}##, ##\beta##.
Relevant Equations: $$\tau_{total} = I_{s} \alpha$$
$$I_{s} = \frac{1}{2}m_{p}R^{2}$$
$$a_{1} = \alpha_{1}R$$

Set up the force equations:
(1) ##mgsin(\beta) - T = ma_{1}##
(2) ##TR = I_{s}\alpha_{1}##

Multiply (1) by ##R## and isolate ##TR##:
##R(mgsin(\beta) - T) = R(ma_{1})##
##mRgsin(\beta) - TR = mRa_{1}##
##TR = mRgsin(\beta) - mRa_{1}##

Plug ##TR## into (2):
##TR = I_{s}\alpha_{1}##
##mRgsin(\beta) - mRa_{1} = I_{s}\alpha_{1}##
##mRgsin(\beta) - mRa_{1} = \frac{1}{2}m_{p}R^{2}\alpha_{1}##

Solve for the ##\alpha_{1}##:
##\alpha_{1} = \frac{mRgsin(\beta) - mRa_{1}}{\frac{1}{2}m_{p}R^{2}}##
(3) ##\alpha_{1} = 2\frac{mgsin(\beta) - ma_{1}}{m_{p}R} = \frac{a_{1}}{R}##

We now use (3) to solve for linear acceleration, ##a_{1}##:
##2\frac{mgsin(\beta) - ma_{1}}{m_{p}R} = \frac{a_{1}}{R}##
##m_{p}a_{1} + 2ma_{1} = 2mgsin(\beta)##
(4) ##a_{1} = \frac{2mgsin(\beta)}{m_{p} + 2m}##

Use (4) to solve the linear kinematics equation for t:
##d = \frac{1}{2} a_{1} t^{2}##
##t = \sqrt{\frac{2d}{a_{1}}}##
$$t = \sqrt{\frac{(m_{p} + 2m)d}{mgsin(\beta)}}$$

I wonder if this solution is correct given that the time to move distance ##d## is independent of the radius ##R## of the pulley. The only intuitive possibility I could think of is a proportional relationship between torque required to achieve a certain angular acceleration and the size of the pulley. Since as the pulley increases as would the leverage and torque required to achieve same angular acceleration. However, I'm not confident in this intuition so I'd love feedback on (a) if my solution is correct and (b) if my intuition explaining the solution is correct. Thank you in advance!
Your solution is correct. A bit inefficient solving for ##\alpha##. Just eliminate it as soon as possible by subbing ##\alpha = \frac{a}{R}## into (1) after subbing (2), And go right to ##a##.
 
  • #3
Thank you!
 
  • #4
giodude said:
I wonder if this solution is correct given that the time to move distance d is independent of the radius R of the pulley. The only intuitive possibility I could think of is a proportional relationship between torque required to achieve a certain angular acceleration and the size of the pulley. Since as the pulley increases as would the leverage and torque required to achieve same angular acceleration.
There's a bit more to it. The MoI rises as the square of R, while both the torque and the distance the mass moves per unit of rotation rise in proportion to R.
It may also seem intuitively wrong because you would expect the pulley's mass to increase too, but it is given as fixed.
 
  • #5
Oh, I think I see. Since the mass of the pulley is fixed, as the size of the pulley increase its really just a purely proportional change because the actual mass isn't increasing (or decreasing) whereas the time dependency is "focused" on the mass of the pulley rather than the shape of it. Which we see by ##m_{p}## being in the solution for time while ##R## is not.
 
  • #6
1696830241692.png
 

Related to MIT OCW 8.01 PS10.6: A Massive Pulley and a Block on an Incline

What is the main objective of MIT OCW 8.01 PS10.6 problem?

The main objective of the problem is to analyze the dynamics of a system involving a massive pulley and a block on an incline, applying principles of Newtonian mechanics to determine the acceleration of the block and the tensions in the ropes.

How do you account for the mass of the pulley in the problem?

The mass of the pulley is taken into account by considering its moment of inertia. The rotational inertia of the pulley affects the overall dynamics of the system, and it must be included in the equations of motion to accurately determine the accelerations and tensions.

What equations are typically used to solve this problem?

The problem is usually solved using Newton's second law for both translational and rotational motion. For the block, F = ma is used, and for the pulley, τ = Iα (where τ is torque, I is moment of inertia, and α is angular acceleration) is applied. Additionally, the kinematic relationship between linear and angular acceleration (a = rα) is used.

How do you find the tension in the rope on either side of the pulley?

The tension in the rope can be found by writing the equations of motion for both the block and the pulley. By solving these equations simultaneously, the tension in the rope on either side of the pulley can be determined. The difference in tension accounts for the torque needed to accelerate the pulley's rotation.

What role does the incline angle play in solving the problem?

The incline angle affects the component of gravitational force acting along the incline, which in turn influences the acceleration of the block. It is crucial to resolve the gravitational force into components parallel and perpendicular to the incline to set up the correct equations of motion.

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