Question

An energetic father stands at the summit of a conical hill as he spins his 24 kg child around on a 5.4 kg cart with a 2.2-m-long rope. The sides of the hill are inclined at 25 ? . He keeps the rope parallel to the ground, and friction is negligible.

What rope tension will allow the cart to spin with the 17 rpm ?

Express your answer to two significant figures and include the appropriate units.

*The answer is not 326.8 N

Answer 1

total mass of the system, m = 24 + 5.4 = 29.4 kg

length of the rope, L = 2.2 m

given, velocity in rpm = 17 rpm = 1.78 rad/s

now, radius of circular track, r = L * cos 25 = 2.2 * cos 25 = 1.994 m

now, linear speed of the system, V =* r

= 1.78 * 1.994 = 3.55 m/s

since the mg*sin 25 will act along the surface of the hill in the downward direction,

Tension force T will be act along the surface in the upward direction direction and also since the cart is moving in the circular path, so a centripiyal force will act as Fc towards the center of the circular path,

so, Fc = T - mg*sin 25

mv²/L = T - mg * sin 25

T = 29.4 * 3.55² /2.2 + 29.4 * 9.81* sin 25 = 290.24 N