- #1
giodude
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- 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!
(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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