Magnitude using Vector: Derivation
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A mass M is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when?
the mass is at the highest point
the wire is horizontal
the mass is at the lowest point
inclined at an angle of 60° from vertical
The equation of motion is given by x(t)=Asinωt+Bcosωt with ω=√Km.
Suppose that at time t=0, the position of mass is x(0) and velocity v(0), then its displacement can also be represented as x(t)=Ccos(ωt−ϕ), where C and ϕ are :
- C=√v(0)2ω2+x(0)2, ϕ=tan−1(v(0)x(0)ω)
- C=√2v(0)2ω2+x(0)2, ϕ=tan−1(x(0)ω2v(0))
- C=√2v(0)2ω2+x(0)2, ϕ=tan−1(x(0)ωv(0))
- C=√2v(0)2ω2+x(0)2, ϕ=tan−1(v(0)x(0)ω)
Repeat the previous exercise if the angle between each pair of springs is 120∘ initially.
- 2π√2mk
- 2π√m2k
- 2π√mk
- 2π√m3k
- About point O, mv0L4=ML212.ω
- About point C, Mv.L4=ML212.ω
- About point A, mv0.3L4=Mv.L2+ML212.ω
- About point B, mv0.3L4=Mv.L+ML212.ω
A person stands on a spring balance at the equator. (a) By what fraction is the balance reading less than his true weight ? (b) If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case ?
(Assume that net torque applied is zero)
- 1:9
- 3:1
- 9:1
- 1:3
A particle of mass m is attached to M via a massless string passing through a hole O in the horizontal table as shown, Mass M is kept stationary whereas mass m is rotating in a circle of radius r with angular speed ω.
Mg=mrω2
Mg>mrω2
Mg≥mrω2
Mg<mrω2
- 2.5 ms−1
- 5.0 ms−1
- 7.5 ms−1
- 10 ms−1
- L2
- L4
- √3L2
- 3L4
- 1:1
- 3:1
- 2:1
- 4:1
- 60√2 kg-wt
- 60 kg-wt
- 120√2 kg-wt
- 600 kg-wt
Assume g=10 ms−2
- 5000√52 N
- 5000√2 N
- 10000√2 N
- 5000√25 N
Each string is 5 m long. Tensions T1 and T2 are experienced by the string when the rod is rotated with 60 rpm.
Take g=10 m/s2 and π2≃10 for calculation.
- Value of T1 is 10.83 N
- Value of T2 is 9.17 N
- Angular velocity of rod is 2π rad/s
- Angular velocity of rod is π rad/s
- 5×10−3 kg/s
- 5×10−5 kg/s
- 10−4 kg/s
- 0.5×10−3 kg/s
[1 kg-wt=10 N]
- 60√2 kg-wt
- 600 kg-wt
- 120√2 kg-wt
- 60 kg-wt
- CA
- AC
- BC
- CB
Choose the correct relation
- N1>N2
- N2>N3
- N3>N1
- N1=N2=N3
- tangential
- centripetal
- no
- tangential
- centripetal
- no
- 3mg2k
- 2mgk
- 3mg4k
- mg4k
- 12.5N
- 25N
- 50N
- 100N
What is centripetal acceleration? Give its formula.
- 1.598
- 398
- 4.598
- 198
A large blade of a helicopter is rotating in a horizontal circle with constant angular velocity. Determine the ratio of acceleration of two particles A and B which are located at a distance of and respectively from the center of the circle.
- 150 N
- 300 N
- 600 N
- 900 N
- −by
- −ay
- 2ay−b
- b−2ay