Question - Circular motion amusement park physics

[clintyip]

Can anyone help me solve this entire Physics question? It's on my exam, and I have worked out some of them, but I have no answers to compare to, so I was wondering if some of you guys can help QUESTION IS:

A new Gravitron Ride is proposed for an amusement park in which the Gravitron will rotate in a vertical circle. The radius of the circular motion is 5.0 metres. The period of revolution is 3.0 seconds at its maximum rotational speed. For a girl of mass 40kg on the ride:-
Calculate:-
(a) The speed of the girl as she revolves
(b) The magnitude of her centripetal acceleration
(c) The maximum force she experiences between herself and the Gravitron

The Gravitron when loaded would have a mass of 1500kg and a radius of gyration of 3.6 metres.
Calculate:
(d) The moment of inertia of the loaded Gravitron
(e) The torque necessary to accelerate the Gravitron to its maximum speed in 10 seconds

So that a smaller motor can be used to drive the ride, it is proposed to accelerate the Gravitron to its maximum speed by using a falling mass attached to a cable wound round a hub of radius 2 metres as shown in the diagram. It is proposed this mass would fall a distance, then be raised by the motor at a slower rate during downtime. i53.tinypic.com/33agux4.png <- Link to diagram

(f) Calculate the mass necessary to accelerate the Gravitron to maximum speed in 10 seconds.
(g) Calculate the distance the mass falls and comment on the feasibility of this arrangementReally need help, it's a matter of what mark I need to get into university.

 

[ehild]

What were your answers and why ?

ehild

 

[clintyip]

What were your answers and why ?

ehild

Okay, here they are:

a) speed = distance/time
= 2 x pi x r / 3 seconds
= 2 x pi x 5 / 3 seconds
speed = 10.47m/s

b) a=v^2/r
a= 10.47^2/5
a= 21.92m/s^2

c) Ftotal = Fcentripetal + Fgravity
= (40x10.47^2/5) + (40x9.8)
= 1269 Newtons
d) I = mr^2
= 1500 x 3.6^2
= 19440kg/m/s^2
e) V = rw
V/r = w
10.47/5 = 2.094m/s

Angular velocity equation form of v = u+at is Wf = Wi + alpha x t
rearranging gives alpha = Wf-Wi/t
Since Wi = 0, equation just becomes alpha = 2.094/10
Therefore alpha = 0.2094m/s^2

Torque = Inertia x Alpha
= 19440 x 0.2094
= 4070 Newton metres

I don't have the answers for (f) and (g)

 

[ehild]

Take care. The girl does not experience centripetal force. She experiences gravity and a force from the seat/seat bell that is the force between herself and the gravitron. The resultant of these forces, Ftotal is the centripetal force. You should write normal force instead. Your numerical results are right.

ehild

 

[clintyip]

Take care. The girl does not experience centripetal force. She experiences gravity and a force from the seat/seat bell that is the force between herself and the gravitron. The resultant of these forces, Ftotal is the centripetal force. You should write normal force instead. Your numerical results are right.

ehild

Would you happen to know how to calculate part f and g?

 

[clintyip]

bump~

 

[ehild]

Would you happen to know how to calculate part f and g?

No, I do not understand the questions, and the figure is not clear. Could you please explain them, and also show your attempt to solve them?


ehild

 

[gneill]

Would you happen to know how to calculate part f and g?

Draw the diagram. Essentially you've got a massive pulley with moment of inertia I, a torque being applied by a cable tension T at radius 2.0m. The tension is provided via a massive weight attached to the cable. You should have seen this sort of problem before.

Hints:
How does the cable tension affect the acceleration of the weight?
How is the angular acceleration of the ride related to the linear acceleration of the weight?

 

[clintyip]

No, I do not understand the questions, and the figure is not clear. Could you please explain them, and also show your attempt to solve them?


ehild

My ATTEMPT of f and g:

f) Fnet = mg - T, where T = tension required to overcome moment of inertia

Torque = F x D
Torque = T x 2

Sub in Torque from previous question, and divide by 2
T = 4070/2
T = 2035N


Combine the following formulas:
Fnet = ma
Fnet = mg-T
a=r(alpha)
Resulting in: mg-T = mr(alpha)
Rearrange formula to make m the subject, making it m = T/g-r(alpha)
sub in all values, and I get that mass = 216.92.

Honestly, I have NO IDEA how to do (g).

 

[ehild]

Rearrange formula to make m the subject, making it m = T/g-r(alpha)
sub in all values, and I get that mass = 216.92.

Yes, but use parentheses in the denominator : m=T/(g-r(alpha)) and do not forget the unit.
As for g, you know the time needed to achieve maximum speed. You know the acceleration of the mass...ehild

 

[clintyip]

Yes, but use parentheses in the denominator : m=T/(g-r(alpha)) and do not forget the unit.
As for g, you know the time needed to achieve maximum speed. You know the acceleration of the mass...ehild

How do I know the acceleration of the mass? o_o

 

[ehild]

The cable moves with the same velocity and aceleration as the rim of the hub when it wounds off. And the mass is at the end of the cable, moving together with it.

ehild

 

[clintyip]

The cable moves with the same velocity and aceleration as the rim of the hub when it wounds off. And the mass is at the end of the cable, moving together with it.

ehild

Ok...So I'm assuming I have to use one of the equations of motion to work out the distance the mass falls. I still don't see where I have an acceleration value, except the acceleration of gravity. But the thing is, the mass isn't free-falling, it's dragging the cable to accelerate the gravitron.

Sorry, I'm not very good at this :(

 

[ehild]

My ATTEMPT of f and g:

f) Fnet = mg - T, where T = tension required to overcome moment of inertia

Torque = F x D
Torque = T x 2

Sub in Torque from previous question, and divide by 2
T = 4070/2
T = 2035N


Combine the following formulas:
Fnet = ma
Fnet = mg-T
a=r(alpha)
Resulting in: mg-T = mr(alpha)
Rearrange formula to make m the subject, making it m = T/g-r(alpha)
sub in all values, and I get that mass = 216.92.

So you have the angular acceleration and the tangential acceleration of the cylinder and it is the same as the acceleration of the mass.


ehild

 

[clintyip]

So, what I got from that is:

a = r(alpha), which is 2x0.2094 = 0.4188m/s^2
S = ut + 1/2 at^2

u = 0, therefore cancel out ut
S = 1/2 at^2
S = 0.5 x 0.4188 x 10^2
S = 20.94metres.

Thing is, would a mass of 216.92kg that falls 20.94metres really be able to accelerate the gravitron of mass 1500kg to max speed in 10 seconds?

 

[ehild]

Well, with friction it would be difficult, but friction is ignored in the problem. And that 21 m is quite much...
I saw a similar thing somewhere at a south beach of England. There was a lift between the beach and the promenade, pulled up by a container filled with water. Reaching the height, the water was let flow out and the lift descended by its own weight raising the container up, which was filled with water there again.

ehild

 

[clintyip]

Well, with friction it would be difficult, but friction is ignored in the problem. And that 21 m is quite much...
I saw a similar thing somewhere at a south beach of England. There was a lift between the beach and the promenade, pulled up by a container filled with water. Reaching the height, the water was let flow out and the lift descended by its own weight raising the container up, which was filled with water there again.

ehild

When I check the validity of the 21 metres, I used Kinetic energy formulas - one for the gravitron itself, and the other one for the gravitational potential energy from the mass.

For Gravitron-
KE = 1/2 I (w)^2
KE = 0.5 x 19440 x 2.094^2
KE = 42620.6 Joules

For Mass-
GPE = mgh
GPE = 216.92 x 9.8 x 21
GPE = 44642.1 Joules

So I GUESS this is valid. :) THANKS A BUNCH!

 

[ehild]

It was a really nice problem, was not it?

ehild

 

[clintyip]

Definitely the most challenging one I've done yet.

Thanks again!