Answer to Question #160896 in Mechanics | Relativity for Malaya M Thorpe

Question

An amusement park ride consists of a rotating circular platform 10.8 m in diameter from which 10 kg seats are suspended at the end of 2.85 m massless chains. When the system rotates, the chains make an angle of 38.2 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2 . θ l d What is the speed of each seat? Answer in units of m/s. 012 (part 2 of 2) 10.0 points If a child of mass 36.8 kg sits in a seat, what is the tension in the chain (for the same angle)? Answer in units of N.

Answer 1

Explanations & Calculations

• The situation could be figured as follows

• As it rotates at some speed (v), the seats along with the chains move outwards the center of rotation just to get supplied the needed centripetal force to move in a circular path.
• The tension of the chains provides this. Its vertical component provides the force needed to keep the seat (or seat+ passenger when loaded) in place without falling down while the horizontal component provides the centripetal force directed towards the center of rotation.

1)

• Apply Newton's second law vertically on the seat

$\qquad\qquad \begin{aligned} \small \uparrow T\cos\theta-mg&= \small 0\\ \small T\cos\theta&= \small mg\\ \small T\cos\theta&= \small mg\cdots(1) \end{aligned}$

• Applying horizontally,

$\qquad\qquad \begin{aligned} \small F&=ma=m\frac{v^2}{r}\\ \small T\sin\theta&= \small m\frac{v^2}{r}\cdots(2)\\ \end{aligned}$

• $\small r =$ the effective distance from the seat (rotating object) to the center of the path

$\qquad\qquad \begin{aligned} \small r&= \small l\sin38.2+\frac{10.8}{2}=2.85\sin38.2+5.4\\ \small &=\small6.769m \end{aligned}$

• By (2)/(1),

$\qquad\qquad \begin{aligned} \small \tan\theta&= \small \frac{v^2}{rg}\\ \small v&= \small \sqrt{\tan38.2\times6.769m\times9.8ms^{-2}}\\ \small v&= \small \bold{6.026\,ms^{-1}} \end{aligned}$

2)

• As the child is on, only the mass of the rotating object (seat+passenger) increases. Using equation (1), the tension can be calculated

$\qquad\qquad \begin{aligned} \small T_1&= \small \frac{Mg}{\cos\theta} \\ &= \small \frac{(10+36.8)kg\times9.8ms^{-2}}{\cos38.2}\\ &= \small \bold{4111.47\,N} \end{aligned}$

• Get the result algebrically first & substitute the numerical values to miimize rounding errors.