Linear speed of sphere as it passes through lowest point

In summary, the conversation discusses the masses and lengths of two spheres and a rod, and the work done by gravity and torque on the system. It is noted that the work done by gravity is equivalent to the work done by torque, and counting it twice would lead to an incorrect answer.
  • #1
dl447342
28
5
Homework Statement
A slender rod is 0.8 m long and has mass 0.12 kg. A small 0.02 kg sphere is welded to one end of the rod, and a small 0.05 kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest. What is the linear speed of the 0.05 kg sphere as it passes through its lowest point?
Relevant Equations
Moment of inertia of slender rod of mass M and length L: ##\frac{1}{12} ML^2##, work done by constant torque through an angular displacement ##\Delta \theta = \tau \Delta\theta##, work-kinetic energy theorem, work done by gravity when an object of mass m falls a vertical distance h is ##mg h##. Kinetic energy = ##\frac{1}2 I \omega^2##.
Let ##m_s = 0.05, m_{s_1} = 0.02, m_r = 0.12, L = 0.8.## be the masses of the two spheres, mass of the rod, and length of the rod. Then the work done by gravity when the rod reaches the vertical position is ##(m_s(L/2) - m_{s_2}(L/2))g## and the kinetic energy equals ##\frac{1}2 (\frac{1}{12} m_rL^2 + \frac{L^2}4 (m_s+m_{s_2}))(\omega_f^2)##, where ##\omega_f## is the final angular speed. If I treat the work done by torque as zero, then I get the correct answer of ##\omega_f = 3.656 rad/s\Rightarrow v_f = 1.46 m/s##, but I thought the work done by torque should be ##\frac{L(m_s-m_{s_2})}2\cdot 9.8##. Why would the torque be zero or is some other quantity different from what I calculated?
 
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  • #2
Realize that gravity provides the torque on this system. So the work done by gravity is the work done by the torque. Don't count it twice.
 
  • #3
Thanks. I was thinking I was counting something twice.
 

1. What is the formula for calculating the linear speed of a sphere as it passes through the lowest point?

The formula for calculating the linear speed of a sphere as it passes through the lowest point is v = √(2gh), where v is the linear speed, g is the acceleration due to gravity, and h is the height of the lowest point.

2. How does the mass of the sphere affect its linear speed as it passes through the lowest point?

The mass of the sphere does not affect its linear speed as it passes through the lowest point. The formula for calculating the linear speed is independent of the mass of the object.

3. Is the linear speed of the sphere constant as it passes through the lowest point?

Yes, the linear speed of the sphere is constant as it passes through the lowest point. This is because the acceleration due to gravity is constant and the height of the lowest point does not change.

4. How does the height of the lowest point affect the linear speed of the sphere?

The height of the lowest point directly affects the linear speed of the sphere. As the height increases, the linear speed also increases. This is because the higher the sphere is, the more potential energy it has, which is converted into kinetic energy as it falls.

5. What is the unit of measurement for linear speed?

The unit of measurement for linear speed is meters per second (m/s). This measures the distance traveled by the sphere in meters divided by the time it takes to travel that distance in seconds.

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