Circular Motion Fun: Calculate Mass Speed at Highest Point

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SUMMARY

The discussion focuses on calculating the speed of a 5.00-kg mass at the highest point of a vertical circular motion, where the mass is attached to a 70.0 cm massless rigid rod. The calculated force exerted by the rod at an angle of 55.0° is 21.9 N. Using the conservation of energy principle, the speed at the highest point is determined to be 3.63 m/s, derived from the equation 1/2mv² = mgh, where h is the height equal to the length of the rod (0.70 m) and g is the acceleration due to gravity (9.8 m/s²).

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Khartoum
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Here's the Question :

A 5.00-kg mass is on the end of a massless rigid rod of length l=70.0cm which pivots freely about the other end. The mass is moving in a vertical circle. When the rod makes an angle of theta=55.0° with the horizontal, the speed of the mass is 1.60 m/s.

a) calculate the magnitude of the force exerted on the mass by the rod.

I already calculated this to be 21.9N and this is correct.

b) (WHERE I NEED HELP) Calculate the speed of the mass at the highest point of the circle.

Now I know I need to write the forces for this point... F = mv²/L - mg

but this equation gives me 2 unknowns, where do I get a value to sub for F (the force exerted by the rod)...or is there some simplifying step I'm missing here...

please help,
thanks.
 
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I assume that the first part specifically asked for the force exerted on the mass by the rod AT 55 degrees above the horizontal?

At the mass's highest point the rod is vertical and its force on the mass is vertical. There is a force of gravity downward on the mass but it is not moving upward or downward so the force of the rod on the mass must be exactly equal to the force of gravity on the mass.
 


To calculate the speed at the highest point of the circle, we can use the conservation of energy principle. At the highest point, all of the kinetic energy of the mass will be converted into potential energy, and we can set these two energies equal to each other.

Kinetic energy at highest point = Potential energy at highest point

1/2mv^2 = mgh

Where v is the speed at the highest point, m is the mass, g is the acceleration due to gravity, and h is the height at the highest point.

We know the height at the highest point is equal to the length of the rod, l=70.0cm. We also know the mass, m=5.00kg. And we can use the acceleration due to gravity, g=9.8m/s^2.

Plugging these values into the equation, we get:

1/2(5.00kg)v^2 = (5.00kg)(9.8m/s^2)(0.70m)

Solving for v, we get:

v = √(9.8m/s^2)(0.70m) = 3.63m/s

Therefore, the speed of the mass at the highest point is 3.63m/s. This makes sense because at the highest point, the mass has no kinetic energy and all of its energy is potential energy. This means the speed will be the slowest at this point.
 

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