InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
An automobile moves on a road54 km/h. The radius of its wheelsis.a speed ofthe moment of inertia of the wheel about iagof rotation is 3 kg-m2. If the vehrest in 15s, the magnitude of average torout itsaxiss broughtuetransmitted by its brakes to wheel is:(1) 2.86 kg-m2/s2 (2) 6.66 kg-m2/s(3) 8.58 kg-mas(4) 10.86 kg-m/s |
|
Answer» Given Conditions⇒ Initial linear velocity(u) = 54 km/hr.= 54× 5/18 [Changing into m/s.]= 15 m/s.Radius of the wheel(r) = 0.35 m. Now, Using the Formula, Initial Angular Velocity (ω₁) = u/rω₁ = 15/0.35ω₁ = 42.86 radian/sec. Also, Final Angular Velocity (ω₂) = 0 Now, Using the Formula,Angular Acceleration (α) = (ω₂ -ω₁)/t= (0 - 42.86)/15= - 2.86 radian/sec² Moment of Inertia about the axis of the rotation(I) = 3 kg-m² ∴ Torque = I×α⇒ Torque = 3× -2.86⇒ Torque = -8.58 Nm. Hence, the average negative torque is -8.58 N-m. thankyou radius is o.45 not 0.35 but no problem I understand concept |
|