Physics Help Forum moment of inertia problem

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 Apr 17th 2019, 06:16 PM #1 Junior Member   Join Date: Apr 2019 Posts: 1 moment of inertia problem 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.
 Apr 18th 2019, 08:49 AM #2 Senior Member   Join Date: Aug 2010 Posts: 434 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 $\displaystyle m_1+ m_2$. topsquark likes this.
Apr 18th 2019, 11:10 AM   #3

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 Originally Posted by austin6 (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
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 Apr 18th 2019, 07:18 PM #4 Junior Member     Join Date: Jan 2019 Posts: 18 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 $\dfrac{1}{2}m_p \cdot a = T_1 - T_2$ $m_1 \cdot a = m_1 g\sin{\theta} - T_1$ $m_2 \cdot a = T_2 - m_2 g$ ------------------------------------------------------- $a\left(\dfrac{1}{2}m_p + m_1 + m_2 \right) = m_1 g \sin{\theta} - m_2 g$ $a = \dfrac{g(m_1\sin{\theta} - m_2)}{\dfrac{1}{2}m_p + m_1 + m_2}$ topsquark likes this.
 Apr 18th 2019, 08:12 PM #5 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,778 Good catch! I glossed over the details of the pulley when I read the problem. -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.

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