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DH (aka Debbie) normally does very good work, I know I am very prone to error, but she took the given diameter as the radius and failed to add the child's mass with the mass of the seat to get the correct tension.

(a) Write an equation for the net centripetal forces acting on a seat. The centripetal force acts horizontally and is caused by the horizontal component of tension in the chain. From Newton’s 2nd law:

∑F(c) = ma(c) = Tsinθ-------->(1)

Here, we use the sine ratio because the angle is measured from the vertical. Vertically, there are two forces acting on the seat. They are the vertical component of tension in the chain and the weight of the seat and child. NSL again:

∑F(y) = 0 = Tcosθ - mg

mg = Tcosθ--------------->(2)

Recalling that sinθ / cosθ = tanθ suggests dividing equation (1) by (2). This has the added benefit of losing the mass and tension:

ma(c) / mg = Tsinθ / Tcosθ

a(c) / g = tanθ

Solving for v (since we know that centripetal acceleration is equal to v²/r):

v²/r / g = tanθ

v² = grtanθ

v = √[grtanθ]

But r is a bit more than just half the diameter of the 8.00m platform because as the platform rotates, the chair swings out a bit. The amount of which is the short leg (opposite side) of a right triangle which has hypotenuse equal to the length of the chain (2.50m). So the additional amount is:

sinθ = opposite side / hypotenuse

opposite side = hypotenuse x sinθ

= 2.50msin31.0°

= 1.29m

Added to the 4.00m radius of the platform gives r as 5.29m, so the speed of a seat is:

= √[(9.80m/s²)(5.29m)tan31.0°]

= 5.58m/s

(b)You should be able to draw this yourself. Make sure that the vectors are pointed in the correct directions, and don’t forget that you have an angle.

(c) You can use either (1) or (2) to find the tension in the chain, (2) is simpler, but note that there might be a very slight difference in the tension between the two equations:

mg = Tcosθ

T = mg/cosθ

= [m(seat) + m(child)]g / cosθ

= (9.90kg + 40.0kg)(9.80m/s²) / cos31.0°

= 571N

Hope this helps.

When the ride rotates, the distance of the rider form the center is 8.00m + 2.50*sin(31) = 9.29m

From a free body diagram about the rider we have T (tension) @31o up and m*g down

Summing vertical forces we get T*cos(31) - m*g = 0 So T = m*g/cos(31)

Now forces in the horizontal ...T*sin(31) This is the centripetal force and = m*v^2/r

So T*sin(31) = m*g*sin(31)/cos(31) = m*v^2/r Note m drops out leaving

v= sqrt(g*r*tan(31)) = sqrt(9.8*9.29*tan(31)) = 7.40m/s

And T = m*g/cos(31) = 40*9.8/cos(31) = 457N