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 $4 H z$ $4 Hz$ to $9 H z$ $9 Hz$ in $3 s$ $3 s$ , what torque was applied to the object?

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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 $4 H z$ $4 Hz$ to $9 H z$ $9 Hz$ in $3 s$ $3 s$ , what torque was applied to the object?

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Answer 1 · 2016-11-07T09:59:16

The torque applied to the object is 53.33 Newton-metre.

Explanation:

As we know torque $τ = I α$ $tau = Ialpha$
So angular acceleration $α$ $alpha$ = (change in angular velocity $ω) / ($ $omega)//($ change in time t)
$\Rightarrow α = (ω_{2} - ω_{1}) / d t$ $=> alpha = (omega_2-omega_1)//dt$
$\Rightarrow α = 5 / 3 \frac{m}{s^{2}}$ $=>alpha=5//3 m/s^2$
And torque now becomes $τ = m r^{2} α [I = m r^{2}]$ $tau= mr^2alpha[I=mr^2]$
By substituting all values we get,
$τ = 2 k g \times {(4 m)}^{2} \times (\frac{5}{3} \frac{m}{s^{2}}) = 53.33$ $tau=2kgxx(4m)^2xx(5/3m/s^2)=53.33$ Newton-metre.

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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 $4 H z$ $4 Hz$ to $9 H z$ $9 Hz$ in $3 s$ $3 s$ , what torque was applied to the object?

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Explanation:

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An object with a mass of 2 kg2kg 2 kg is traveling in a circular path of a radius of 4 m4m4 m. If the object's angular velocity changes from 4 Hz4Hz 4 Hz to 9 Hz9Hz 9 Hz in 3 s3s 3 s, what torque was applied to the object?

1 Answer

Explanation:

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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 $4 H z$ $4 Hz$ to $9 H z$ $9 Hz$ in $3 s$ $3 s$ , what torque was applied to the object?