 # 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. ringearV 2020-10-26 Answered
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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Given
${m}_{1}=1kg$ and ${m}_{2}=2kg$
Distance between them is 1metre.
$x=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}}$
$⇒x=\frac{1\cdot 0+2\cdot 1}{1+2}=\frac{2}{3}=0.67m$
Moment of inertia is
$I={m}_{1}\cdot {x}_{1}^{2}+{m}_{2}\cdot {x}_{2}^{2}=1\cdot {\left(0.67\right)}^{2}+2\cdot {\left(0.332\right)}^{2}$
$I=0.4489+0.2178=0.6667=0.67k\frac{g}{{m}^{2}}.$
As per Angular Impulse
$\tau ×t={L}_{1}-{L}_{2}$
$\tau =\frac{{L}_{1}-0}{t}$
$\tau =\frac{{I}_{1}{\omega }_{1}}{t}$
$\tau =\frac{0.67\cdot 2\cdot 3.14\cdot 0.33}{5}=0.28N\cdot m$
$\omega =20r±=\frac{20}{60}=0.33rps$