Derive an expression for maximum speed of a vehicle moving along a horizontal circular track.
The maximum speed of a vehicle moving along a horizontal circular track occurs when the centripetal force is equal to the frictional force. The centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{r} \] where \( m \) is the mass of the vehicle, \( v \) is the velocity, and \( r \) is the radius of the track. The frictional force \( F_f \) is given by:
\[ F_f = \mu mg \] where \( \mu \) is the coefficient of friction and \( g \) is the acceleration due to gravity. Equating the two forces:
\[ \frac{mv^2}{r} = \mu mg \] Solving for \( v \), we get the maximum speed:
\[ v_{{max}} = \sqrt{\mu g r} \]
A body of mass $100 \;g$ is moving in a circular path of radius $2\; m$ on a vertical plane as shown in the figure. The velocity of the body at point A is $10 m/s.$ The ratio of its kinetic energies at point B and C is: (Take acceleration due to gravity as $10 m/s^2$)
If a body is performing uniform circular motion with velocity \( v \) and radius \( R \), then identify the true statements from the following:
A. Its velocity \( v \) is constant.
B. Acceleration is always directed towards the centre and its magnitude is \( a = \frac{v^2}{R} \).
C. Angular momentum is constant in magnitude but its direction keeps changing.
D. Angular velocity of the body \( = \frac{v}{R} \).
Choose the most appropriate answer from the options given below.
The slope of the tangent to the curve \( x = \sin\theta \) and \( y = \cos 2\theta \) at \( \theta = \frac{\pi}{6} \) is ___________.
Solve the following L.P.P. by graphical method:
Maximize:
\[ z = 10x + 25y. \] Subject to: \[ 0 \leq x \leq 3, \quad 0 \leq y \leq 3, \quad x + y \leq 5. \]