What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency $9 H z$ $9 Hz$ over $2 s$ $2 s$ ?

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What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency $9 H z$ $9 Hz$ over $2 s$ $2 s$ ?

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Answer 1 · 2018-04-03T13:25:47

The torque for the rod rotating about the center is $= 75.49 N m$ $=75.49Nm$
The torque for the rod rotating about one end is $= 301.69 N m$ $=301.69Nm$

Explanation:

The torque is the rate of change of angular momentum

$τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I \frac{d ω}{d t} = I α$ $tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt=Ialpha$

The mass of the rod is $m = 2 k g$ $m=2kg$

The length of the rod is $L = 4 m$ $L=4m$

The moment of inertia of a rod, rotating about the center is

$I = \frac{1}{12} \cdot m L^{2}$ $I=1/12*mL^2$

$= \frac{1}{12} \cdot 2 \cdot 4^{2} = 2.67 k g m^{2}$ $=1/12*2*4^2= 2.67 kgm^2$

The rate of change of angular velocity is

$α = \frac{d ω}{d t} = \frac{9}{2} \cdot 2 π$ $alpha=(domega)/dt=(9)/2*2pi$

$= (9 π) r a d s^{- 2}$ $=(9pi) rads^(-2)$

So the torque is $τ = 2.67 \cdot (9 π) N m = 75.49 N m$ $tau=2.67*(9pi) Nm=75.49Nm$

The moment of inertia of a rod, rotating about one end is

$I = \frac{1}{3} \cdot m L^{2}$ $I=1/3*mL^2$

$= \frac{1}{3} \cdot 2 \cdot 4^{2} = 10.67 k g m^{2}$ $=1/3*2*4^2=10.67kgm^2$

So,

The torque is $τ = 10.67 \cdot (9 π) = 301.69 N m$ $tau=10.67*(9pi)=301.69Nm$

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What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency $9 H z$ $9 Hz$ over $2 s$ $2 s$ ?

1 Answer

Explanation:

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What torque would have to be applied to a rod with a length of $4 m$ $4 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency $9 H z$ $9 Hz$ over $2 s$ $2 s$ ?