Circular Motion and Universal Gravitation

[wvlaxxer81]

Ok, well my Honors Physics teacher doesn't really like to "teach" but rather give us a list of equations, and then throw a packet at us and tells us to do it. Last semester I got a C and I know that I could do much better if I was actually explained the information and how and when to apply the certain equations in certain situations.

Anyway, I just wanted to put down these problems we received in our most recent packet to see if I'm totally lost or not. Any feedback is greatly appreciated.1. A 25.0 Kg child is on a merry-go-round moving with a speed of 1.25 m/s at a distance of 11.0 m from the center. Calculate: a) the centripetal acceleration of the child, b) the Fc on the child, c) the coefficient of friction, u, necessary for the child to stay on without holding on. (Realize this may not be a reasonable number!)

My attempt:

a) ac=v^2/r ac = (1.25m/s)^2/11m Centripetal Acceleration = 0.14 m/s^2

b) Fc = m(ac) Fc = 25 Kg x (0.14 m/s^2) Fc = 3.5N

c) Fc = f(riction) f = uN 3.5N = u(25Kg x 9.8 m/s^2) u = 0.014N2. A ball of mass 0.335 Kg is swung at uniform rate in a vertical circle of radius 85.0 cm with a speed of 3.25m/s. Calculate the tension in the string when the ball is: a) at the top of its path, b) at the bottom of its path.

My attempt:

Fg = m(ag) , ac = v^2/r , Fc = mac , Fc = Fg - Fn (at the top) , Fc = Fn - Fg (at the bottom)

ac = 3.25^2/.85 (changed cm to m) = 12.4 m/s^2
Fc = .335 Kg(12.4m/s^2) = 4.154N
Fg = .335(9.8m/s^2) = 3.28N

a) Fc = Fg - Fn 4.154N = 3.28N - Fn Fn = -0.874N

b) Fc = Fn - Fg 4.154N = Fn - 3.28N Fn = 7.434N3. A 1000 Kg car rounds a curve of radius 65.0 m. If the car is traveling at 90.0 km/hr, what friction force is required for them to stay on the road?

My attempt:

Equations: ac = V^2/r , Fc = m(ac) , f = uN , N = m(ag) , Fc = f

ac = (90 m/s)^2 / 65m = 124.61 m/s^2 , Fc = 1000 Kg(124.61 m/s^2) = 124610N

N = 1000 kg x 9.8m/s^2 = 9800N , 124610N = u9800N u = 12.72N

*I don't know if I should convert the 90 km/hr to m/hr (or 65m to km) because the answer seemed very unreasonable*4. What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to Earth is 1/25 of its value at the surface?

My Attempt:

Equations: (1/25)Fg = 0.392 m/s^2 , ac = v^2/r

9.8m/s^2 = V^2/6.37x10^6m (Earth's radius) V^2 = 6242600 m/s --> take square root of

each side so: V = 2498.52 m/s.

0.392 m/s^2 = 2498.52^2/r r = 15,925,005.59m5. A jet pilot takes his aircraft in a vertical loop. If the jet is moving at a speed of 700 Km/hr at the bottom of the loop, determine the minimum radius of the loop if the acceleration at the lowest point doesn't exceed 6.0 g's.

My Attempt:

9.8m/s^2 x 6 g's = 58.8m/s^2 , a = v^2/r , 58.8m/s^2 = (700 km/hr)^2/r r = 8333.3m6. A Geosynchronous satellite is one that stays above the same spot on the earth. What would its velocity have to be if it is 3200 Km above the surface of the earth.

Equations: Delta X = 2(3.14)r , V = delta X/delta T , Radius of Earth = 6.37x10^6m

Delta X = 2(3.14...)(6370000 m + 3200000 m) = 60130083.39 m
V= 60130083.39 m / 24 hours = 2505420.14 m/hr or 2505.42 Km/hr7. What is the maximum speed a 1000 Kg car can round a turn of a radius 85.0m on a flat road if the coefficient of friction between the tires and road is 0.6? Does this speed depend on the mass of the car?

My Attempt:

Equations: f = uN , a = v^2/r

f = u(m(ag)) or m(a) = u (m(ag)) so the masses cancel so a = u(ag) a = .6(9.8m/s^2)

a = 5.88 m/s^2 , v^2/85 = 5.88 , V^2 = 499.8 , square root both sides and V = 22.36 m/s

And then I said that the mass of the car doesn't matter since the masses will cancel out in the beginning of the equation anyway.
Thank your for any help you can give. I realize this is a little long and I'm sorry if it is too long but even just input on a question or two will be really helpful.

Thank you.

 

[Redbelly98]

Welcome to Physics Forums!

1. A 25.0 Kg child is on a merry-go-round moving with a speed of 1.25 m/s at a distance of 11.0 m from the center. Calculate: a) the centripetal acceleration of the child, b) the Fc on the child, c) the coefficient of friction, u, necessary for the child to stay on without holding on. (Realize this may not be a reasonable number!)

My attempt:

a) ac=v^2/r ac = (1.25m/s)^2/11m Centripetal Acceleration = 0.14 m/s^2

b) Fc = m(ac) Fc = 25 Kg x (0.14 m/s^2) Fc = 3.5N

c) Fc = f(riction) f = uN 3.5N = u(25Kg x 9.8 m/s^2) u = 0.014N

a & b are correct.
c is the correct number, but the wrong units.

2. A ball of mass 0.335 Kg is swung at uniform rate in a vertical circle of radius 85.0 cm with a speed of 3.25m/s. Calculate the tension in the string when the ball is: a) at the top of its path, b) at the bottom of its path.

My attempt:

Fg = m(ag) , ac = v^2/r , Fc = mac , Fc = Fg - Fn (at the top) , Fc = Fn - Fg (at the bottom)

ac = 3.25^2/.85 (changed cm to m) = 12.4 m/s^2
Fc = .335 Kg(12.4m/s^2) = 4.154N
Fg = .335(9.8m/s^2) = 3.28N

a) Fc = Fg - Fn 4.154N = 3.28N - Fn Fn = -0.874N

b) Fc = Fn - Fg 4.154N = Fn - 3.28N Fn = 7.434N

Essentially correct, however:
a) Fc = Fg +[/color] Fn
(each of these three force terms is in the same direction, downward)

Also, note that when a problem asks for "the tension", you just need to specify it's magnitude (i.e. absolute value). The direction of the force is generally clear from the context -- strings and ropes always pull, they never push.

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I suggest posting separate questions in separate threads. That will get different people helping you. That's better than having each person saying to themself "Too bad I don't have time to answer 7 questions", and ignoring you.

Also, if it turns out there are lots of problems with your answers, it would get pretty confusing discussing all 7 questions at the same time.

Thanks for showing the equations and your work on solving the problems, that does make it easier to help you