What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $2 H z$ $2 Hz$ over $6 s$ $6 s$ ?

Question

What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $2 H z$ $2 Hz$ over $6 s$ $6 s$ ?

1 Answer

Impact of this question

1324 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-23T14:46:12

The torque (for the rod rotating about the center) is $= 8.73 N m$ $=8.73Nm$
The torque (for the rod rotating about one end) is $= 34.91 N m$ $=34.91Nm$

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

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

$= \frac{1}{12} \cdot 2 \cdot 5^{2} = \frac{25}{6} k g m^{2}$ $=1/12*2*5^2= 25/6 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{2}{6} \cdot 2 π$ $(domega)/dt=(2)/6*2pi$

$= (\frac{2}{3} π) r a d s^{- 2}$ $=(2/3pi) rads^(-2)$

So the torque is $τ = \frac{25}{6} \cdot (\frac{2}{3} π) N m = \frac{25}{9} π N m = 8.73 N m$ $tau=25/6*(2/3pi) Nm=25/9piNm=8.73Nm$

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

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

$= \frac{1}{3} \cdot 2 \cdot 5^{2} = \frac{50}{3} k g m^{2}$ $=1/3*2*5^2=50/3kgm^2$

So,

The torque is $τ = \frac{50}{3} \cdot (\frac{2}{3} π) = \frac{100}{9} π = 34.91 N m$ $tau=50/3*(2/3pi)=100/9pi=34.91Nm$

Science

Math

Humanities

... and beyond

What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $2 H z$ $2 Hz$ over $6 s$ $6 s$ ?

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 5m5 m and a mass of 2kg2 kg to change its horizontal spin by a frequency of 2Hz2 Hz over 6s6 s?

1 Answer

Explanation:

Related questions

Impact of this question

What torque would have to be applied to a rod with a length of $5 m$ $5 m$ and a mass of $2 k g$ $2 kg$ to change its horizontal spin by a frequency of $2 H z$ $2 Hz$ over $6 s$ $6 s$ ?