A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid, massless rod. The rod is rotating cw about its center of mass at 20 rpm. What torque will bring the balls to a halt in 5.0 s? Answer in Nm.

Question

Answer 1

Given

m_{1} = 1 k g

and

m_{2} = 2 k g

Distance between them is 1metre.

x = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}

\Rightarrow x = \frac{1 \cdot 0 + 2 \cdot 1}{1 + 2} = \frac{2}{3} = 0.67 m

Moment of inertia is

I = m_{1} \cdot x_{1}^{2} + m_{2} \cdot x_{2}^{2} = 1 \cdot {(0.67)}^{2} + 2 \cdot {(0.332)}^{2}

I = 0.4489 + 0.2178 = 0.6667 = 0.67 k \frac{g}{m^{2}} .

As per Angular Impulse

τ \times t = L_{1} - L_{2}

τ = \frac{L_{1} - 0}{t}

τ = \frac{I_{1} ω_{1}}{t}

τ = \frac{0.67 \cdot 2 \cdot 3.14 \cdot 0.33}{5} = 0.28 N \cdot m

ω = 20 r \pm = \frac{20}{60} = 0.33 r p s

A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid, massless rod. The rod is rotating cw about its center of mass at 20 rpm. What torque will bring the balls to a halt in 5.0 s? Answer in Nm.

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