What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $1 k g$ $1 kg$ to change its horizontal spin by a frequency of $2 H z$ $2 Hz$ over $2 s$ $2 s$ ?

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What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $1 k g$ $1 kg$ to change its horizontal spin by a frequency of $2 H z$ $2 Hz$ over $2 s$ $2 s$ ?

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Answer 1 · 2017-03-29T09:11:55

The torque, for the rod rotating about the center is $= 4.71 N m$ $=4.71Nm$
The torque, for the rod rotating about one end is $= 18.85 N m$ $=18.85Nm$

Explanation:

The torque is the rate of change of angular momentum
$τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I \frac{d ω}{d t}$ $tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

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

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

$= \frac{1}{12} \cdot 1 \cdot 3^{2} = 0.75 k g m^{2}$ $=1/12*1*3^2=0.75 kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $τ = 0.75 \cdot (2 π) N m = 4.71 N m$ $tau=0.75*(2pi) Nm=4.71Nm$

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

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

$= \frac{1}{3} \cdot 1 \cdot 3^{2} = 3$ $=1/3*1*3^2=3$

So,

The torque is $τ = 3 \cdot (2 π) = 6 π = 18.85 N m$ $tau=3*(2pi)=6pi=18.85Nm$

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What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $1 k g$ $1 kg$ to change its horizontal spin by a frequency of $2 H z$ $2 Hz$ 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 3m3 m and a mass of 1kg1 kg to change its horizontal spin by a frequency of 2Hz2 Hz over 2s2 s?

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What torque would have to be applied to a rod with a length of $3 m$ $3 m$ and a mass of $1 k g$ $1 kg$ to change its horizontal spin by a frequency of $2 H z$ $2 Hz$ over $2 s$ $2 s$ ?