An object with a mass of $2 k g$ $2 kg$ is traveling in a circular path of a radius of $2 m$ $2 m$ . If the object's angular velocity changes from $3 H z$ $3 Hz$ to $8 H z$ $8 Hz$ in $1 s$ $1 s$ , what torque was applied to the object?

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Answer 1 · 2018-03-28T14:45:38

The torque is $= 251.3 N m$ $=251.3Nm$

The torque is the rate of change of angular momentum

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

where $I$ $I$ is the moment of inertia

For the object, $I = m r^{2}$ $I=mr^2$

The mass is $m = 2 k g$ $m=2 kg$

The radius of the path is $r = 2 m$ $r=2m$

So the moment of inertia is

, $I = 2 \cdot {(2)}^{2} = 8 k g m^{2}$ $I=2*(2)^2=8kgm^2$

The angular acceleration is

$α = \frac{2 π f_{1} - 2 π f_{0}}{t} = (16 - 6) \frac{π}{1} = 10 π r a d s^{- 2}$ $alpha=(2pif_1-2pif_0)/t=(16-6)pi/1=10pirads^-2$

So,

The torque is

$τ = I α = 8 \cdot 10 π = 251.3 m$ $tau=Ialpha=8*10pi=251.3m$

An object with a mass of 2 kg2kg 2 kg is traveling in a circular path of a radius of 2 m2m2 m. If the object's angular velocity changes from 3 Hz3Hz 3 Hz to 8 Hz8Hz 8 Hz in 1 s1s 1 s, what torque was applied to the object?