Circular motion and speed of rotation

[kikifast4u]

Homework Statement

There's a 'chair-o-plane' with a radius of 4m and the seats are connected using 3m chains.
a) Calculate the speed of rotation at which the chains make an angle of 35 to the vertical
b) What is the tension in the chain if a rider of 90kg is on the swing at the speed above?

Homework Equations

Newton's laws, Centripetal force formula, Trigonometric formulas

The Attempt at a Solution

I did a) and found a reasonable value of about 11RPM.
However I have a problem with b).
I get to equations, T'*sin(35)=g(M+m) and T'*cos(35)=a(M+m), where T' is the tension and M and m are the masses of the rider and the chair. Since I have a=g*tan(35) (from a) ), the two equations above are equivalent and I therefore have one equations with two unknowns. Am I missing something? How can I get m in order to find T'?

 

[tiny-tim]

hi kikifast4u!

There's a 'chair-o-plane' with a radius of 4m and the seats are connected using 3m chains.
a) Calculate the speed of rotation at which the chains make an angle of 35 to the vertical
b) What is the tension in the chain if a rider of 90kg is on the swing at the speed above?
…
I get to equations, T'*sin(35)=g(M+m) and T'*cos(35)=a(M+m), where T' is the tension and M and m are the masses of the rider and the chair. Since I have a=g*tan(35) (from a) ), the two equations above are equivalent and I therefore have one equations with two unknowns. Am I missing something? How can I get m in order to find T'?

i think you're supposed to assume that the mass m of the chair is negligible compared with the mass M of the rider, ie that m = 0

 

[kikifast4u]

I do hope so. Otherwise I don't think I have enough data to get a numerical value. Thanks!

 

[Pr.George]

kikifast4u,
This is professor George, i believe by signing the Own Work declaration you agree not to use the help of any external agency to solve your Assignments.
This is viewed by the university as a serious offence.
I had a look at your paper and i already got your name, your assignment won't be corrected or graded unless you head to the ETO and confess you used external help.
Failure to do so will result in more than losing grades.
Thank you.

 

[rwisz]

I would like to clarify that the circular motion and speed of rotation in this scenario are related to the concept of centripetal force. The chair-o-plane is moving in a circular motion, and the speed of rotation is determined by the radius of the ride and the angle at which the chains are connected. This speed can be calculated using the formula v = ωr, where v is the linear speed, ω is the angular velocity, and r is the radius.

In order to find the tension in the chains, we need to consider the forces acting on the rider. There are two forces at play here – the weight of the rider (mg) and the centripetal force (Fc). The centripetal force is provided by the tension in the chains and is directed towards the center of the circular motion.

To find the tension, we can use the equation Fc = mv^2/r, where m is the mass of the rider and v is the linear speed calculated in part a). We can also write Fc = Tcos(35), where T is the tension in the chains and cos(35) is the component of the tension in the direction of the centripetal force.

Setting these two equations equal to each other, we get Tcos(35) = mv^2/r. This gives us the tension in the chains, T = (mv^2)/(rcos(35)).

Now, to find the mass of the rider (m), we can use the equation Fg = mg, where Fg is the weight of the rider and g is the acceleration due to gravity. We can also write Fg = Tsin(35), where sin(35) is the component of the tension in the direction opposite to the weight of the rider.

Setting these two equations equal to each other, we get Tsin(35) = mg. This gives us the mass of the rider, m = (Tsin(35))/g.

Substituting this value of m into the equation for tension, we get T = (mv^2)/(rcos(35)) = [(Tsin(35))/g]*v^2/(rcos(35)).

Simplifying, we get T = (Tsin(35)v^2)/(grcos(35)). Solving for T, we get T = (v^2)/(gcos(35)). Plugging in the values for v and g