Question

A belt is run over two drums. The larger drum has weight 4 lbs and a radius of gyration of 1.25 inches while the smaller drum has weight 2.7 lbs and a radius of gyration of 0.75 inches. The tension from the smaller drum is held constant at 6 lbs. If it is known that the speed of the belt is 11 ft/s after 0.16 s, what is the tension between the drums?

Answer 1

269 lb

Step-by-step explanation:

We first find the tangential acceleration, a on the drums

a = Δv/Δt since the speed of the belt is 11 ft/s after 0.16 s, Δv = 11 ft/s and Δt = 0.16 s

a = Δv/Δt = 11 ft/s ÷ 0.16 s = 68.75 ft/s²

Since torque τ = Tk = Iα where I = moment of inertia of larger drum = Mk² where m = mass of larger drum = 4 lbs, k = radius of gyration of larger drum = 1.25 inches, T = tension due to larger drum and α = angular acceleration of larger drum.

So, T = Iα/k = Mk²α/k = Mαk = Ma (since a = αk )

T = 4 lbs × 68.75 ft/s² = 275 lb

The tension due to the smaller drum is T' = 6 lb .

So the net tension in the belt is T'' = T - T' = 275 lb - 6 lb = 269 lb