Question

A 4-kg sphere is attached to two wires AC and BC. If the sphere rotates about its axis at constant speed, compute the minimum and maximum velocity if the tension in either of the wires is not to exceed 35 N. Solution: Vmin

Answer 1

vmin = 2.816 m/s

vmax = 3.1517 m/s

Step-by-step explanation:

We apply Newton's Second Law as follows

∑Fy = m*g ⇒ TAC*Cos 30° + TBC*Cos 45° = m*g (1)

∑Fx = m*v²/R ⇒ TAC*Sin 30° + TBC*Sin 45° = m*v²/R (2)

Subtract Eq. (1) from Eq. (2). Since Sin 45° = Cos 45°, we obtain

TAC*(Cos 30°-Sin 30°) = (m*g) - (m*v²/R) (3)

Multiply Eq. (1) by Sin 30°, Eq. (2) by Cos 30°, and subtract:

TBC*(Sin 30°*Cos 45°-Cos 30°*Sin 45°) = Sin 30°*m*g - Cos 30°*(m*v²/R)

- TBC*Sin 15° = Sin 30°*m*g - Cos 30°*(m*v²/R) (4)

Making

TAC = 35 N, m = 4 Kg, R = 1.2 m, g = 9.81 m/s² in Eq. (3), we find the value v₁ of v for which TAC = 35 N:

35 N*(Cos 30°-Sin 30°) = (4 Kg*9.81 m/s²) - (4 Kg*v₁²/1.2 m)

⇒ 12.8109 = 39.24 - 3.33*v₁² ⇒ v₁ = 2.816 m/s

We have TAC ≤ 35 N for v ≥ v₁, that is, for v ≥ 2.816 m/s

Making

TBC = 35 N, m = 4 Kg, R = 1.2 m, g = 9.81 m/s² in Eq. (4), we find the value v₂ of v for which TBC = 35 N:

- 35 N*Sin 15° = Sin 30°*4 Kg*9.81 m/s² - Cos 30°*(4 Kg*v₂²/1.2 m)

⇒ - 9.059 N = 19.62 N - 2.887 Kg*v₂²/m

⇒ v₂ = 3.1517 m/s

We have TBC ≤ 35 N for v ≤ v₂ , that is, for v ≤ 3.1517 m/s

Combining the results obtained, we conclude that the range of allowable value is 2.816 m/s ≤ v ≤ 3.1517 m/s