How to Calculate Speed in Circular Motion on a Hemisphere Surface?

In summary, the conversation discusses a particle at rest on a smooth fixed hemisphere with a horizontal base and center O. The particle is then displaced and moves on the surface of the hemisphere. At a point P with angle AOP = α, the particle has speed v. The task is to find an expression for v in terms of a, g, and α. The solution is to equate the total energy at point A (PE+KE = amg) with the total energy at point P (PE+KE = (0.5m(v^2)) + xmg). However, there was an error in calculating the initial potential energy, which should be PE=mg*(
  • #1
furor celtica
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0

Homework Statement



A particle is at rest at the apex A of a smooth fixed hemisphere whose base is horizontal. The hemisphere has centre O and radius a. The particle is then displaced very slightly from rest and moves on the surface of the hemisphere. At the point P on the surface where angle AOP = α the particle has speed v. Find an expression for v in terms of a, g and α.

Homework Equations





The Attempt at a Solution



So I’ve worked like this:
Total energy at A = PE + KE = amg + 0
Total energy at P = PE + KE = (0.5m(v^2)) + xmg
x = a – y
(cos α)/y = (sin90)/a => y = a(cos α)
=> x = a – (a(cos α)) = a (1 - cos α)
=> Total energy at P = PE + KE = (0.5m(v^2)) + amg (1 - cos α)

Therefore (Total energy at A) = (Total energy at P) gives
amg = (0.5m(v^2)) + amg (1 - cos α)
ag = 0.5(v^2) + ag (1 – cos α)
ag – ag (1 – cos α) = 0.5(v^2)
2ag (1 – 1 + cos α) = v^2
v = sqrt (2ag (cos α))

However, the correct answer is v = sqrt (2ag (1 - cos α))

Where’s the problem?
 

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  • #2
furor celtica said:

The Attempt at a Solution



So I’ve worked like this:
Total energy at A = PE + KE = amg + 0
Total energy at P = PE + KE = (0.5m(v^2)) + xmg
You calculated the initial potential energy with respect to the base of the semiphere. PE=mg*(height above the base). The height is y in your picture.

ehild
 
  • #3
thanks
 

1. What is circular motion?

Circular motion is the movement of an object along a circular path. This means that the object is constantly changing direction, but maintains a constant distance from a central point.

2. How is speed calculated in circular motion?

In circular motion, speed is calculated by dividing the distance traveled by the object by the time it takes to travel that distance. This is known as the average speed of the object.

3. Can an object have a constant speed in circular motion?

Yes, an object can have a constant speed in circular motion. This is known as uniform circular motion, where the speed remains constant while the direction of motion changes.

4. How does the radius affect the speed in circular motion?

The radius of the circular path affects the speed in circular motion. The larger the radius, the greater the distance the object has to travel, resulting in a higher speed. Similarly, a smaller radius will result in a lower speed.

5. What is centripetal acceleration in circular motion?

Centripetal acceleration is the acceleration of an object moving in a circular path. It always points towards the center of the circle and is directly proportional to the square of the object's speed and inversely proportional to the radius of the circle.

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