What torque would have to be applied to a rod with a length of $8 m$ $8 m$ and a mass of $8 k g$ $8 kg$ to change its horizontal spin by a frequency $18 H z$ $18 Hz$ over $9 s$ $9 s$ ?

Question

What torque would have to be applied to a rod with a length of $8 m$ $8 m$ and a mass of $8 k g$ $8 kg$ to change its horizontal spin by a frequency $18 H z$ $18 Hz$ over $9 s$ $9 s$ ?

1 Answer

Impact of this question

1371 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-05T14:25:48

The torque for the rod rotating about the center is $= 536.2 N m$ $=536.2Nm$
The torque for the rod rotating about one end is $= 2144.7 N m$ $=2144.7Nm$

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

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

$= \frac{1}{12} \cdot 8 \cdot 8^{2} = 42.7 k g m^{2}$ $=1/12*8*8^2= 42.7 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{18}{9} \cdot 2 π$ $(domega)/dt=(18)/9*2pi$

$= (4 π) r a d s^{- 2}$ $=(4pi) rads^(-2)$

So the torque is $τ = 42.7 \cdot (4 π) N m = 536.2 N m$ $tau=42.7*(4pi) Nm=536.2Nm$

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

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

$= \frac{1}{3} \cdot 8 \cdot 8^{2} = 170.7 k g m^{2}$ $=1/3*8*8^2=170.7kgm^2$

So,

The torque is $τ = 170.7 \cdot (4 π) = 2144.7 N m$ $tau=170.7*(4pi)=2144.7Nm$

Science

Math

Humanities

... and beyond

What torque would have to be applied to a rod with a length of $8 m$ $8 m$ and a mass of $8 k g$ $8 kg$ to change its horizontal spin by a frequency $18 H z$ $18 Hz$ over $9 s$ $9 s$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What torque would have to be applied to a rod with a length of 8 m8m8 m and a mass of 8 kg8kg8 kg to change its horizontal spin by a frequency 18 Hz18Hz18 Hz over 9 s9s9 s?

1 Answer

Explanation:

Related questions

Impact of this question

What torque would have to be applied to a rod with a length of $8 m$ $8 m$ and a mass of $8 k g$ $8 kg$ to change its horizontal spin by a frequency $18 H z$ $18 Hz$ over $9 s$ $9 s$ ?