-->

Units And Measurement

Question
CBSEENPH11020651

An automobile moves on a road with a speed of 54 km h-1. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m2 . If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its breaks to the wheel is

  • 6.66 kg m2s-2

  • 8.58 kg m2s-2

  • 10.86 kg m2s-2

  • 2.86 kg m2s-2

Solution

A.

6.66 kg m2s-2

As velocity of an automobile vehicle
straight v space equals space 54 space km divided by straight h space equals space 54 space straight x space 5 over 18 space equals space 15 space straight m divided by straight s
Angular space velocity space of space straight a space vehicle comma space straight v space equals space straight omega subscript straight o straight r

rightwards double arrow space straight omega subscript straight o space equals space straight v over straight R space equals space fraction numerator 15 over denominator 0.45 end fraction space equals space 100 over 3 space rad divided by straight s
So comma space angular space acceleration space of space an space antomobile
straight alpha space equals space fraction numerator increment straight omega over denominator straight t end fraction space equals space fraction numerator straight omega subscript straight t minus straight omega subscript straight o over denominator straight t end fraction space equals space fraction numerator 0 minus begin display style 100 over 3 end style over denominator 15 end fraction space equals space fraction numerator negative 100 over denominator 45 end fraction space rad divided by straight s squared
Thus comma space average space torque space transmitted space by space its space brakes space to space wheel
torque space equals space Iα
rightwards double arrow space 3 space straight x space 100 over 45 space equals space 6.66 space kg space straight m squared straight s to the power of negative 2 end exponent