A point on the periphery of rotating disc has its acceleration vector making on angle 30° with velocity vector then the ratio of magnitude of centripetal acceleration to tangential acceleration is :
A block on a stationary horizontal table with increasing speed in a circle is seen from an inertial frame of reference. The angle between net force on the block and velocity vector is :
When net acceleration makes acute angle with velocity, speed of the particle will increase. As tangential accleration is positive, speed will increase.
A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle θ1; in the next 4 sec, it rotates through an additional angle θ2. The ratio of θ2/θ1 is :
For Circular motion angle traversed is θ = ω t + ½ α t²
so, θ1 =½α(2)² = 2α
and θ1 + θ2 = ½α(6)² = 18 α ⇒ θ2= 16 α
so θ2 /θ1= 8
At t = 0 a wheel is rotating at 50 rad/sec. A motor gives it a constant angular acceleration of 5 rad/sec2 until it reaches 100 rad/sec the motor is disconnected how many revolutions are completed at t = 20 sec
ω = ω 0 + α.t
⇒ 100 =50 + 5 × t
⇒ 50 = 5.t
⇒ t = 10 sec.
Angle traversed during acceleration= ½.α.t²= ½×5×(10)²= 250 rad
Angle traversed with constant angular speed = ω.t= 100 × 10 = 1000 rad
Total angle traversed = 1250 rad
Number of Revolutions = 1250 /2π= 625 /π
Find ratio of Normal reaction of blocks. At point A & point B respectively if their masses are same and velocities are 2 m/s & 4 m/s respectively :
Centrifugal Force act radially outwards,
So, N A = mg - mv²/R = m(10) - m(2)²/2= 8m
So, N B = mg + mv²/R = m(10) +m(4)²/2= 18m
N A : N B = 4 : 9
Two stones of masses m and 2 m are whirled in horizontal circles, the heavier one in a radius r/2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is :
The kinetic energy \((K)\) of particle moving along a circle of radius \(R\) depends upon the distance covered \(S\) and is given by \(K = aS\) where \(a\) is a constant. Then the centripetal force acting on the particle is:
Kinetic energy is \(K = \frac{1}{2}mv^2 = aS\). Centripetal force is \(F_c = \frac{mv^2}{R}\) ( Since \(mv^2 = 2aS\), we get \(F_c = \frac{2aS}{R}\).
A particle of mass \(m\) is tied to a string of length \(L\) and rotated in vertical circle about other end with critical speed so that it is just able to complete the vertical loop. Then tension in string, when string is at horizontal position will be:
To just complete the vertical loop, the velocity at the bottom is \(\sqrt{5gL}\). By energy conservation, the velocity at the horizontal position is \(v = \sqrt{3gL}\). The tension at this point is \(T = \frac{mv^2}{L} = 3mg\).
