moment of inertia problem

Apr 2019
1
0
For the system shown in the figure below, m1 = 7.0 kg, m2 = 2.0 kg, θ = 31°, and the radius and mass of the pulley are 0.10 m and 0.10 kg, respectively.
https://imgur.com/a/VYiRKTo
(a) What is the acceleration of the masses? (Neglect friction and the string's mass.)
Known Formulas:
Torque= Inertia x Angular Accel.
I=(MR^2)/2
F=ma

(b) The tension acting on m1 is different from the tension acting on m2. Why?


After attempting to try a force diagram I am stumped still. Any advice in the correct direction would be appreciated.
 
Aug 2010
434
174
I don't see that this has anything to do with "moment of inertia". The object on the slope has mass \(\displaystyle m_1\) so gravitational force (weight) \(\displaystyle m_1g\) downward. The slope is at angle \(\displaystyle \theta\) so the component of weight perpendicular to the slope \(\displaystyle m_1g cos(\theta)\) and the component parallel to the slope is \(\displaystyle m_1g sin(\theta)\).

The component perpendicular to the slope is canceled by the slope itself. It is the component parallel to the slope that is supported by the other mass. Its weight is \(\displaystyle m_2g\) so the net force, to the left is \(\displaystyle m_1gsin(\theta)-
m_2g\). The acceleration of the two objects is that divided by the total mass \(\displaystyle m_1+ m_2\).
 
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topsquark

Forum Staff
Apr 2008
3,020
637
On the dance floor, baby!
(b) The tension acting on m1 is different from the tension acting on m2. Why?
This statement is wrong. If we are dealing with an ideal string the tension on both ends of the string must be the same. (Do a FBD on the string itself.)

Are you sure this is the right diagram for this problem? Like HallsofIvy mentioned, this is not a question that has anything to do with rotational motion.

-Dan
 
Jan 2019
55
41
The tensions are different because a torque is generated about the pulley.

$I \cdot \alpha = (T_1 - T_2) \cdot r$

assuming the pulley is a uniform disk

$\dfrac{1}{2}m_p r^2 \cdot \dfrac{a}{r} = (T_1 - T_2) \cdot r$

$\dfrac{1}{2}m_p \cdot a = T_1 - T_2$

sum the rotational equation with the two translational equations term for term

$\dfrac{1}{2}m_p \cdot a = T_1 - T_2$

$m_1 \cdot a = m_1 g\sin{\theta} - T_1$

$m_2 \cdot a = T_2 - m_2 g$
-------------------------------------------------------
$a\left(\dfrac{1}{2}m_p + m_1 + m_2 \right) = m_1 g \sin{\theta} - m_2 g$

$a = \dfrac{g(m_1\sin{\theta} - m_2)}{\dfrac{1}{2}m_p + m_1 + m_2}$
 
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topsquark

Forum Staff
Apr 2008
3,020
637
On the dance floor, baby!
Good catch! I glossed over the details of the pulley when I read the problem.

-Dan