Radius of Curvature
Trending Questions
- v2bRg
- vRgb
- v2Rg
- v2bR
- u2cos2θ2g
- √3u2cos2θ2g
- u2cos2θg
- √3u2cos2θg
- 254√10 m
- 352√10 m
- 252√5 m
- 254√5 m
What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle θ2 with the horizontal ?
- 10 N
- 20 N
- 30 N
- 40 N
If a cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4 m, then coefficient of friction between the cycle tyres and road is
0.61
0.71
0.51
0.41
- 316 m
- 100√5 m
- 100√10 m
- 300 m
- 1440 N
- 320 N
- 640 N
- 720 N
- 0.075 kg/m3
- 0.133 kg/m3
- 0.75 kg/m3
- 1.33 kg/m3
- a2
- 12a
- 4a
- 32a
A train is moving towards north. At one place it turns towards north-east, here we observe that
The radius of curvature of outer rail will be greater than that of the inner rail
The radius of the inner rail will be greater than that of the outer rail
The radius of curvature of the outer and inner rails will be the same
The radius of curvature of one of the rails will be greater
- 160 N
- 320 N
- 1280 N
- 640 N
- 10√2 m/s
- 250√5 m/s
- 10√5 m/s
- 250√2 m/s
[Takeg=10 m/s2]
- 10 m/s
- 12 m/s
- 5 m/s
- 0
- (v0cosθ)2/g
- v20/g
- v20cosθ/g
- v20/gcosθ
- y=x2+constant
- y2=x+constant
- xy=constant
- y2=x2+constant
- 250√5 m/s
- 10√2 m/s
- 10√5 m/s
- 250√2 m/s
- 25 m
- 15 m
- 12.5 m
- 20 m
- 10 N
- 20 N
- 30 N
- 40 N
- √3mv38g
- 3mv38g
- 3mv316g
- √3mv316g
- u2cos22θgcos3θ
- u2cos22θgcos2θ
- u2cos2θgcos3θ
- u2cos32θgcos2θ
- 90o
- 30o
- 14o
- 16o
- RB>RC>RA
- RC>RB>RA
- RB>RA>RC
- RA>RB>RC
A particle is projected at angle θ with horizontal with velocity V0
Find
(i) tangential and normal acceleration of the particle at t = 0 and at highest point of its trajectory.
(ii) radius of curvature at t = 0 and highest point.
at
At highest point
at
At highest point
1
1.5
2
2.33
- 2 s
- 4 s
- 6 s
- 8 s
- 20cm
- 50cm
- 100cm
- 40cm