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A stone tied to a string is rotated in a vertical plane. If mass of the stone is m, the length of the string is R and the linear speed of the stone is v when the stone is at its lowest point, then the tension in the string at the lowest point will be :
(centripetal force at lowest point)
The correct answer is:
A ring of radius R lies in vertical plane. A bead of mass m can move along the ring without friction. Initially the bead is at rest at the bottom most point on ring. The minimum constant horizontal speed v with which the ring must be pulled such that the bead completes the vertical circle :
In the frame of ring (inertial w.r.t. earth), the initial velocity of the bead is v at the lowest position.
The condition for bead to complete the vertical circle is, its speed at top position
From conservation of energy
When the road is dry and coefficient of friction is μ the maximum speed of a car in a circular path is 10ms^{–1}. If the road becomes wet and coefficient of friction become μ/4, what is the maximum speed permitted?
The correct answer is: 5ms^{–1}
A car of mass M is traveling on a horizontal circular path of radius r. At an instant its speed is v and tangential acceleration is a :
friction force on car
which is greater than
∴ it is not less than a/g for safe turn.
The correct answer is: The magnitude of frictional force on the car is greater than
A body is tied up by a string of length l and rotated in vertical circle at minimum speed. When it reaches at highest point string breaks and body moves on a parabolic path in presence of gravity according to figure. In the plane of point A, value of horizontal range AC will be :
The correct answer is:
The maximum velocity at the lowest point, so that the string just slack at the highest point in a vertical circle of radius l.
The correct answer is:
A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is released from position A and leaves the track at B. The radius of curvature of its trajectory when it just leaves the track at B is:
By energy conservation between A and B
Now, radius of curvature
The correct answer is: R/2
A particle of mass m begins to slide down on a fixed smooth sphere from the top. What is its tangential acceleration when it breaks off the sphere?
At loose contact N = 0
from energy conservation
from (1) & (2)
tangential acceleration
The correct answer is:
A particle is projected horizontally from the top of a tower with a velocity v_{0}. If v be its velocity at any instant, then from radius of curvature of the path of the particle at that instant is directly proportional to :
As we know :
(centripetal acceleration)
From figure :
The correct answer is: v^{3}
The breaking tension of a string is 10N. A particle of mass 0.1kg tied to it is rotated along a horizontal circle of radius 0.5metre. The maximum speed with which the particle can be rotated without breaking the string is :
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