Calculating torque on a pendulum

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SUMMARY

The discussion focuses on calculating the torque on a pendulum consisting of a rod with a mass of 1 kg and a solid sphere with a mass of 0.5 kg. The torque formula used is τ = -rFsinθ, where the total force is derived from the combined mass of the rod and sphere multiplied by gravitational acceleration. The calculated torque at a 30-degree angle is -9.6 Nm, which led to a discrepancy with the book's answer key. The user resolved the issue by considering the torque exerted by the weight of the rod, aided by a reference to a Walter Lewin video for further clarification.

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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.
 
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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
 

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