Centripetal acceleration of the stone

[hytuoc]

Would someone please show me how to do these problems below. Thanks.
1) An Earth satellite moves in a circular orbit 640km above Earth's surface w/ a period of 98.0min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

2) a boy whirls a stone in a horizontal circle of radius 1.5m and at height 2.0 above level ground. the string breaks, and the stone flies off horizontally and strikes the ground after traveling a horizontal distance of 10m. What is the magnitude of the centripetal acceleration of the stone while in circular motion?

 

[Leong]

#1
Use v=\frac{2\pi r}{T} with
v= the speed of the satellite
r= the radius of the circular orbit of the satellite
T= the period taken by the satellite to make one revolution

Use a=\frac{v^2}{r} with
a= the magnitude of the centripedal acceleration of the satellite
v= the speed of the satellite
r= the radius of the circular orbit of the satellite

#2
Use a=\frac{v^2}{r} with
a= the magnitude of the centripedal acceleration of the stone
v= the speed of the stone when the string breaks
r= the radius of circular path of the stone
Manipulate the formula to obtain v which is the horizontal initial velocity of the stone when the string breaks. The rest is about 2D projectile motion.

 

[wais]

Sure, I'd be happy to help you with these problems! Let's start with the first one about the Earth satellite. To find the speed of the satellite, we can use the formula v = 2πr/T, where v is the speed, r is the radius of the orbit, and T is the period. Plugging in the values given, we get:

v = 2π(640km)/(98.0min) = 13.1km/min

To find the magnitude of the centripetal acceleration, we can use the formula a = v^2/r, where a is the centripetal acceleration and v is the speed we just calculated. Plugging in the values, we get:

a = (13.1km/min)^2/(640km) = 0.268km/min^2

For the second problem, we can use the formula a = v^2/r to find the centripetal acceleration of the stone while it is still in circular motion. We know the radius of the circle (1.5m) and the distance the stone travels (10m), so we just need to find the speed of the stone. We can use the formula v = d/t, where v is the speed, d is the distance, and t is the time. Since the stone travels 10m in a horizontal direction, we can assume that it takes the same time to travel that distance as it did to complete one full circle. So we can use the period given in the problem, which is 98.0min. Converting that to seconds, we get 98.0min x 60s/min = 5880s.

Now we can plug in the values to find the speed:

v = (10m)/(5880s) = 0.0017m/s

Finally, we can plug this speed and the radius into the formula for centripetal acceleration:

a = (0.0017m/s)^2/(1.5m) = 0.000002m/s^2

I hope this helps you understand how to approach these types of problems! Let me know if you have any further questions.