Question

I need help with a homework question: An amusement park ride consists of a rotating circular platform 9.87 m in diameter from which 10kg seats are suspended at the end of 2.63 m massless chains. When the system rotates, the chains makes an angle of 33.9 degrees with the vertical. The acceleration of gravity is 9.8 m/s^2. What is the speed of each seat in m/s?

Answer 1

Use Newton's Second Law of Motion to find the centripetal acceleration of each seat. Next, use the relation between the centripetal acceleration, the speed of the seats and the radius of the circular trajectory to find the speed of the seats.

There are two forces acting on each seat: the tension of the chains and the weight of the seat.

The vertical component of the tension of the seats must balance their weight, and the horizontal component of the tension of the chains is responsible for the centripetal acceleration of the seats.

Draw a free body diagram to visualize the situation:

Since the seats have a mass of 10kg, their weight must be equal to 98 Newtons.

On the other hand, we can find T using the vertical component of the tension:

`\begin{gathered} T_y=98N=T\cdot\cos (33.9º) \\ \Rightarrow T=(98N)/(\cos (33.9º)) \end{gathered}`

Replace the expression for T to find the horizontal component of the tension:

`\begin{gathered} T_x=T*\sin (33.9º) \\ =(98N)/(\cos(33.9º))*\sin (33.9º) \\ =98N*\tan (33.9º) \\ =65.85N \end{gathered}`

Using Newton's Second Law of Motion, find the centripetal acceleration, taking into account that the net force is equal to the horizontal component of the tension:

When an object moves in a circular trajectory with radius r and tangential speed v, the centripetal acceleration is given by:

Isolate v from the equation:

`v=\sqrt[]{a_c\cdot r}`

To find the radius of the trajectory, notice that since the 2.63m chains ar tilted forming an angle of 33.9º with the vertical, they add a horizontal distance to the radius of the platform given by:

`\Delta r=(2.63m)*\sin (33.9º)=1.467m`

Since the diameter of the platform is 9.87m, the radius of the platform is half that distance. Then, the radius of the trajectory is:

`r=(9.87m)/(2)+1.467m=6.4m`

Replace the values for the centripetal acceleration and the radius of the trajectory to find the speed of the seats:

`v=\sqrt[]{6.4m*6.585(m)/(s^2)}=6.5(m)/(s)`

Therefore, the speed of each seat is 6.5 meters per second.