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Icetray
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Hi guys,
I have this question below but I don't have the answer to it and I'm not sure if what I am doing is right. I attempted the solution by balancing the forces as I am not as clear on how to use the energy method yet.
Is my solution/answer correct? (:
Thanks guys!Homework Statement
Consider the system shown in Figure 10.42 with m1 = 20.0kg, m2 = 12.5kg, R(radius of pully)=0.2m and mass of uniform pulley M=5.00kg. Object m2 is resting on the floor and m1 is 4.00 meters above the floor when it is released from rest. The pully axis is frictionless. Calculate the time for m1 to hit the floor.
Relevant equations and attempt at solution
M1:
T[itex]_{1}[/itex] = m[itex]_{1}[/itex]g - m[itex]_{1}[/itex]a
T[itex]_{1}[/itex] = 196.2 - 20a
M2:
T[itex]_{2}[/itex] = m[itex]_{2}[/itex]g + m[itex]_{2}[/itex]a
T[itex]_{2}[/itex] = 122.625 + 12.5a
Pully:
T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = I[itex]\alpha[/itex]
T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = (0.5)(MR[itex]^{2}[/itex])(a/R)
T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = 0.5(5.00)(2.00)a
Subbing in T[itex]_{1}[/itex] and T[itex]_{2}[/itex] I get a = 2.23m/s[itex]^{2}[/itex]
Then I use Xf = Xi + Ut +0.5at[itex]^{2}[/itex] to get t and get 1.89s
Is this right?
I have this question below but I don't have the answer to it and I'm not sure if what I am doing is right. I attempted the solution by balancing the forces as I am not as clear on how to use the energy method yet.
Is my solution/answer correct? (:
Thanks guys!Homework Statement
Consider the system shown in Figure 10.42 with m1 = 20.0kg, m2 = 12.5kg, R(radius of pully)=0.2m and mass of uniform pulley M=5.00kg. Object m2 is resting on the floor and m1 is 4.00 meters above the floor when it is released from rest. The pully axis is frictionless. Calculate the time for m1 to hit the floor.
Relevant equations and attempt at solution
M1:
T[itex]_{1}[/itex] = m[itex]_{1}[/itex]g - m[itex]_{1}[/itex]a
T[itex]_{1}[/itex] = 196.2 - 20a
M2:
T[itex]_{2}[/itex] = m[itex]_{2}[/itex]g + m[itex]_{2}[/itex]a
T[itex]_{2}[/itex] = 122.625 + 12.5a
Pully:
T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = I[itex]\alpha[/itex]
T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = (0.5)(MR[itex]^{2}[/itex])(a/R)
T[itex]_{1}[/itex] - T[itex]_{2}[/itex] = 0.5(5.00)(2.00)a
Subbing in T[itex]_{1}[/itex] and T[itex]_{2}[/itex] I get a = 2.23m/s[itex]^{2}[/itex]
Then I use Xf = Xi + Ut +0.5at[itex]^{2}[/itex] to get t and get 1.89s
Is this right?
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