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An object with a mass of # 2 kg# is traveling in a circular path of a radius of #7 m#. If the object's angular velocity changes from # 7 Hz# to # 9 Hz# in # 3 s#, what torque was applied to the object?

Answer 1

Charlotte Aaron

410.7 Nm

#"Torque = Rate of change of Angular Momentum"#

#= (L_f - L_i) / t#

#= (I ω_f - I ω_i) / t# ………[∵ Angular Momentum = #I ω#]

#= (I(ω_f - ω_i))/t#

#= (2piI(f_f - f_i))/t# …………[#∵ ω = 2pif#]

#=(2pixxmr^2(f_f - f_i))/t#

#=(2xx22/7 × 2kg × (7 m)^2 × (9Hz - 7Hz))/(3s)#

#=410.7# Nm

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Answer 2

Lily Adams

Torque = Moment of Inertia × Angular Acceleration Moment of Inertia (I) for a point mass rotating at a radius (r) = m × r^2 Angular Acceleration (α) = (Final Angular Velocity - Initial Angular Velocity) / Time

Calculate I, then α, and use the torque formula.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.