Verifying Torque and Motion Problem Solution

[Hockeystar]

Homework Statement

All info is in attachment

Homework Equations

All info is in attachment

The Attempt at a Solution

I'm asking if someone can verify whether or not my work is correct. I'm doubting whether I solved the problem correctly or not.

 

[graphene]

looks alright

 

[inky]

Homework Statement

All info is in attachment

Homework Equations

All info is in attachment

The Attempt at a Solution

I'm asking if someone can verify whether or not my work is correct. I'm doubting whether I solved the problem correctly or not.

I wonder table is smooth or not.

 

[hikaru1221]

You got the wrong acceleration. Even if friction is neglected, the pulley has to be taken into account, as it's not massless. Here is my solution (assume that the table is frictionless):
_ For the 2-kg block:
m_2g - T_2 = m_2a
T_2 = 20 - 2a
_ For the 4-kg block:
T_4 = m_4a
T_4 = 4a
_ For the pulley:
T_2r - T_4r - M_f = I.\alpha = \frac{1}{2}m_pr^2\alpha = \frac{1}{2}m_pra
T_2 - T_4 - 2.5 = 1.5a
Thus:
20 - 2a - 4a - 2.5 = 1.5a
a = \frac{7}{3} m/s^2
_ The torque on the pulley:
M_p = I.\alpha = \frac{1}{2}m_pr^2\alpha = \frac{1}{2}m_pra
M_p = 0.7 Nm

Oh they ask for the torque on the pulley, which means this torque includes the friction's torque.

 

[inky]

You got the wrong acceleration. Even if friction is neglected, the pulley has to be taken into account, as it's not massless. Here is my solution (assume that the table is frictionless):
_ For the 2-kg block:
m_2g - T_2 = m_2a
T_2 = 20 - 2a
_ For the 4-kg block:
T_4 = m_4a
T_4 = 4a
_ For the pulley:
T_2r - T_4r - M_f = I.\alpha = \frac{1}{2}m_pr^2\alpha = \frac{1}{2}m_pra
T_2 - T_4 - 2.5 = 1.5a
Thus:
20 - 2a - 4a - 2.5 = 1.5a
a = \frac{7}{3} m/s^2
_ The torque on the pulley:
M_p = I.\alpha = \frac{1}{2}m_pr^2\alpha = \frac{1}{2}m_pra
M_p = 0.7 Nm

Oh they ask for the torque on the pulley, which means this torque includes the friction's torque.

You mean tension on the string are not the same for 2 bodies. We consider same tension. According to hockeystar, you should take g= 9.8 ms-2.

 

[hikaru1221]

You mean tension on the string are not the same for 2 bodies. We consider same tension. According to hockeystar, you should take g= 9.8 ms-2.

It is impossible for the tensions to equal. Since the pulley is not massless, if the tensions equal, their torques on the pulley cancel each other, plus the frictional torque which prevents the pulley to rotate, the pulley will no way rotate.
If the pulley is massless, its moment of inertia is zero. From the equation: torque = (moment of inertia) x (angular acceleration), we shall see that torque = 0.

Sorry for not noticing g=9.8 m/s^2 :) So all to do is just recalculating.

 

[inky]

It is impossible for the tensions to equal. Since the pulley is not massless, if the tensions equal, their torques on the pulley cancel each other, plus the frictional torque which prevents the pulley to rotate, the pulley will no way rotate.
If the pulley is massless, its moment of inertia is zero. From the equation: torque = (moment of inertia) x (angular acceleration), we shall see that torque = 0.

Sorry for not noticing g=9.8 m/s^2 :) So all to do is just recalculating.

Thanks . I got a lot of knowledge.