An object with a mass of $2 k g$ $2 kg$ is traveling in a circular path of a radius of $4 m$ $4 m$ . If the object's angular velocity changes from $1 H z$ $1 Hz$ to $5 H z$ $5 Hz$ in $2 s$ $2 s$ , what torque was applied to the object?

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Answer 1 · 2018-02-14T06:38:00

The torque is $= 402.1 N m$ $=402.1Nm$

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

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

The radius of the path is $r = 4 m$ $r=4m$

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

So, $I = 2 \cdot {(4)}^{2} = 32 k g m^{2}$ $I=2*(4)^2=32kgm^2$

The initial angular velocity is

$ω_{0} = 2 π f_{0} = 2 \cdot π \cdot 1 = 2 π r a d s^{- 1}$ $omega_0=2pif_0=2*pi*1=2pirads^-1$

The final angular velocity is

$ω_{1} = 2 π f_{1} = 2 \cdot π \cdot 5 = 10 π r a d s^{- 1}$ $omega_1=2pif_1=2*pi*5=10pirads^-1$

The time is $t = 2 s$ $t=2s$

The angular acceleration is

$α = \frac{ω_{1} - ω_{0}}{t} = \frac{10 π - 2 π}{2} = 4 π r a d s^{- 2}$ $alpha=(omega_1-omega_0)/t=(10pi-2pi)/2=4pirads^-2$

So,

The torque is $τ = 32 \cdot 4 π = 402.1 N m$ $tau= 32*4pi=402.1Nm$

An object with a mass of 2kg 2 kg is traveling in a circular path of a radius of 4m4 m. If the object's angular velocity changes from 1Hz 1 Hz to 5Hz 5 Hz in 2s 2 s, what torque was applied to the object?