Friction and Horizontal Circular Motion Problems
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- 10√3 m/s
- 20 m/s
- 10√5 m/s
- 10 m/s
A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small wooden plate is kept on the seat with its plane perpendicular to the radius of the circular road (figure 7-E4). A small block of mass 100 g is kept on the seat which rests against the plate. The friction coefficient between the block and the plate is μ=0.58. (a) Find the normal contact force exerted by the plate on the block. (b) The plate is slowly turned so that the angle between the normal to the plate and the radius of the road slowly increases. Find the angle at which the block will just start sliding on the plate.
A car moving on a horizontal road may be thrown out of the road in taking a turn
By the gravitational force
Due to lack of sufficient centripetal force
Due to rolling frictional force between tyre and road
Due to the reaction of the ground
- the same throughout the motion.
- minimum at the highest position of the circular path.
- minimum at the lowest position of the circular path.
- minimum when the rope is in the horizontal position.
- 0.3
- 0.15
- 0.4
- 0.2
[Take g=10 m/s2]
- 0.8
- 0.55
- 0.7
- 0.9
- 10 m/s
- 8 m/s
- 13 m/s
- 6 m/s
- 10√15 m/s
- 3√35 m/s
- 50 m/s
- 5√15 m/s
- mg[1−(ωR−2v)ωg−v2Rg]
- mg[1−2(ωR−v)ωg−v2Rg]
- mg[1−(ωR−2v)ωg+v2Rg]
- mg[1−2(ωR−v)ωg+v2Rg]
(Take, g=10 m/s2)
- tan−1(23)
- tan−1(35)
- tan−1(25)
- tan−1(14)
- 20 m/s
- 25 m/s
- 15 m/s
- 14 m/s
- 16000 N
- 8000√3 N
- 4000 N
- 4000√3 N
- tan−1(110)
- tan−1(10)
- tan−1(120)
- tan−1(1100)
- Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
- Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
- Assertion is correct but Reason is incorrect
- Assertion is incorrect but Reason is correct
- 40 m/s
- 23.72 m/s
- 25.72 m/s
- 50 m/s
- tan−1(110)
- tan−1(10)
- tan−1(120)
- tan−1(1100)
- mω2R
- mω2R2π
- mω2R2
- mω2Rπ
1) Inner wheels leave the ground first.
2) Outer wheels leave the ground first.
- √gR2μs+tanθ1−μstanθ
- √gRμs+tanθ1−μstanθ
- √gRμs+tanθ1−μstanθ
- √gR2μs+tanθ1−μstanθ
[Take g=10 m/s2]
- 10 m/s
- 8 m/s
- 13 m/s
- 6 m/s
- 10√3 m/s
- 10 m/s
- 20 m/s
- 10√5 m/s
- 0.5
- 0.6
- 0.9
- 1.1
- 2π√R/g
- 12π√R/g
- π√R/g
- 12π√Rg
- 2.94
- 0.58
- 0.34
- 0.23
- 10 m/s
- 8 m/s
- 13 m/s
- 6 m/s
- ω2r
- ωr
- vω
- vr
- 8.7 J
- 10.7 J
- 7.8 J
- 12.7 J