Go Back   Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old Jul 2nd 2017, 07:10 PM   #1
Junior Member
 
Join Date: Jul 2017
Posts: 1
Rotational Motion Question!

Hello!

So this is the question:

A uniform rigid rod with mass Mr = 2.7 kg, length L = 3.1 m rotates in the vertical xy plane about a frictionless pivot through its center. Two point-like particles m1 and m2, with masses m1 = 6.7 kg and m2 = 1.6 kg, are attached at the ends of the rod. What is the magnitude of the angular acceleration of the system when the rod makes an angle of 51.1 degrees with the horizontal? (m2 is 51.1 degrees above the horizontal, and m1 is 51.1 degrees below the horizontal).

My efforts:

I began by calculating inertia, like this:

I = mL2/12 = (2.7kg)(3.1m)2/12=2.16225 kgm2

Then I calculated the gravitational forces from m1 and m2:

m1g=(6.7kg)(9.8m/s2)=65.66N
m2g=(1.6kg)(9.8m/s2)=15.68N

Then I determined the component of these forces which act perpendicular to the lever arm:

65.66N*cos51.1=41.23N
15.68N*cos51.1=9.84N

Then I calculated the net torque/moment (clockwise negative, counterclockwise positive):

Radius = 1/2(3.1m)=1.55m
Torque = (1.55m41.23N)-(1.55m9.84N) = 48.65Nm

Then I calculated the angular acceleration:

Angular acceleration = Torque/Inertia
Angular acceleration = 48.65Nm/2.16kg*m2
Angular acceleration = 22.52 rad/s2

I have been told my answer is incorrect through an online system where I am able to check my answers, however I'm not sure where I went wrong. If anyone would be able to spot an error, or guide me in the right direction, it would be greatly appreciated!!
shmoop is offline   Reply With Quote
Old Jul 3rd 2017, 10:03 AM   #2
Senior Member
 
skeeter's Avatar
 
Join Date: Aug 2008
Posts: 110
Looks like you omitted the rotational inertia of the endpoint masses

$\alpha = \dfrac{\tau_{net}}{I}$

$\tau_{net} = m_1g\cos{\theta} \cdot \frac{L}{2} - m_2g\cos{\theta} \cdot \frac{L}{2} \approx 48.65 \, N~m$

$I_{sys} = m_1 \left(\frac{L}{2}\right)^2 + m_2 \left(\frac{L}{2}\right)^2 + \frac{m_r L^2}{12} \approx 22.1 \, kg~m^2$

$\alpha \approx 2.2 \, rad/s^2$
skeeter is offline   Reply With Quote
Reply

  Physics Help Forum > High School and Pre-University Physics Help > Kinematics and Dynamics

Tags
motion, question, rotational



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Rotational Motion Sheldon Kinematics and Dynamics 2 Nov 5th 2012 09:29 AM
rotational motion tanaki Kinematics and Dynamics 2 Sep 22nd 2009 03:33 AM
Rotational Motion sake Advanced Mechanics 1 Mar 24th 2009 10:58 PM
rotational motion hriday Advanced Mechanics 1 Jan 7th 2009 06:05 PM
Rotational Motion noppawit Kinematics and Dynamics 1 Sep 4th 2008 08:41 AM


Facebook Twitter Google+ RSS Feed