A stone of mass 1 kg tied to one end of a string 1.0 m long is revolved in a horizontal circle at the rate of 10/π revolution per second. Calculate the tension of the string ?
Tension force in the string will provide required centripetal force
T = mω²r= 1× (10/π ×2π)²×1= 400 N
A spot light S rotates in a horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light P moves along the wall at a distance of 3 m. The velocity of spot P when θ = 45° is :-
A store of mass 5 kg is tied to a string of length 10 m is whirled round in a horizontal circle. What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N :
Tension force will provide required centripetal force so,
T = mv²/r
⇒ 200 = 5 v²/10 ⇒ v² = 400 ⇒ v =20 m/s
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
A cyclist is moving with speeding up itself in a circular track then work done by net force on cyclist will be :
As the speed of cyclist is increasing positive tangential acceleration is acting on the object. so net acceleration makes an acute angle with velocity.
As, net acceleration and net force will have same direction, angle between velocity and net force is acute. Work done will be positive.
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 :
A particle moves in a circle of radius 5 m with constant speed and time period 2π s. The acceleration of the particle is :
As speed of the particle is constant only centripetal acceleration will act on the object.
Centripetal Acceleration = ω²R
a c = (2π/2π)× 5= 5 m/s²
Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v0, then the ratio of tensions in the three sections of the string is :
T c = mω²(3l)= 3mω²
T B = mω²(2l) + T c = 5mω²l
T A = mω²(l) + T B = 6mω²l
So,
T C : T B : T A = 3:5:6
The maximum speed of car with which it can go around a level road of radius 10 m is (coefficient of friction between the road and tyre is 0.5)
(g = 9.8 m/s²)
The maximum speed on a level road is limited by static friction: u=v²/rg
Given μs=0.5, g=9.8m/s², and r=10m.
Substituting these values: V²max=(0.5)(9.8)(10)= 49 = 7m/s.
Therefore, the maximum speed is 7 m/s.