A wheel of a cycle is rotating at a speed of \(10\) revolutions per second. After \(20Sec\), it is rotating with a speed of \(20\) revolutions per second. Calculate: the angle through which wheel is turned in \(20Sec\).
Ans: \(\theta=600\pi rad\).
A constant torque acts on a wheel and turned it from rest through \(400rad.\) in \(10Sec\). Find: (i) Angular acceleration of wheel. (ii) If, some torque continues to act what will be the angular velocity of the wheel after \(20Sec\) from start?
Ans: \(\omega=160\frac{rad.}{Sec}\).
A constant torque is acting on a wheel. If, starting from rest, the wheel makes \(n-rotation\) in time \(ts\) then show that the angular acceleration is given by \(\alpha=\left(\frac{4\pi n}{t^{2}}\right)\frac{rad}{Sec}\).
A fly-wheel of mass \(1MT\) and radius of \(1m\) is rotating at a rate of \(420r.p.m\) . What will be the constant retarding torque required to stop the wheel in \(14\) rotations. Assuming; that mass is uniformly distributed on the rim of the wheel only.
Ans: \(\tau=-11\times10^{3}N-m\) .
Th radius of the wheel of a vehicle is \(50cm\) . The vehicle is accelerated from rest by an angular acceleration of \(2\frac{rad}{s^{2}}\) in \(20s\). How much distance is covered by wheel in this time interval and what will be its linear velocity?
Ans: \(S=200m\) and \(v=20\frac{m}{s}\).
The motor of an engine is rotating about its axis with an angular speed of \(100r.p.m\). It comes to rest in \(15s\) after being switched off. Assuming constant angular deacceleration, calculate the number of revolutions made by it before coming to rest.
