Calculating torque on a pendulum

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The discussion revolves around calculating the torque on a pendulum consisting of a rod and a sphere. The initial calculation yielded a torque of -9.6, but the user questioned the accuracy due to discrepancies with the book's answer key. A suggestion was made to consider the torque exerted by the weight of the rod, which led to a breakthrough in understanding. A reference to a Walter Lewin video was provided as a helpful resource for others facing similar issues. Ultimately, the discussion highlights the importance of accurately accounting for all forces in torque calculations.
I_Try_Math
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Homework Statement
A pendulum consists of a rod of mass 1 kg and
length 1 m connected to a pivot with a solid
sphere attached at the other end with mass 0.5
kg and radius 30 cm. What is the torque about
the pivot when the pendulum makes an angle of
30 with respect to the vertical?
Relevant Equations
## \tau = -rFsin\theta ##
7-16-q2.jpg

Let ##m_{r}=1## kg be the mass of the rod and ##m_{s}=0.5## kg be the mass of the sphere.
## \tau = -rFsin\theta ##
## = -r([m_{r}+m_{s}]g)sin\theta ##
## =-1.3(1.5)(9.8)sin30 ##
## \tau = -9.6 ##
My book's answer key disagrees and my initial thoughts are that maybe the mass in my calculation is incorrect. Any help is appreciated.
 
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I_Try_Math said:
Homework Statement: A pendulum consists of a rod of mass 1 kg and
length 1 m connected to a pivot with a solid
sphere attached at the other end with mass 0.5
kg and radius 30 cm. What is the torque about
the pivot when the pendulum makes an angle of
30 with respect to the vertical?
Relevant Equations: ## \tau = -rFsin\theta ##

View attachment 348431
Let ##m_{r}=1## kg be the mass of the rod and ##m_{s}=0.5## kg be the mass of the sphere.
## \tau = -rFsin\theta ##
## = -r([m_{r}+m_{s}]g)sin\theta ##
## =-1.3(1.5)(9.8)sin30 ##
## \tau = -9.6 ##
My book's answer key disagrees and my initial thoughts are that maybe the mass in my calculation is incorrect. Any help is appreciated.
Think carefully about the torque exerted by the weight of the rod.
 
Well I finally figured this one out after that hint. If anyone comes across this in the future here's a Walter Lewin video that might help
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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