Motion in a Plane Easy Questions and Answers | Page - 10

Motion in a Plane

An electric fan has blades of length 30 cm measured from the axis of rotation. If the fan is rotating at 120 rpm, the acceleration of a point on the tip of the blade is

1. 1600 ms^–2
1. 47.4 ms^–2
1. 23.7 ms^–2
1. 50.55 ms^–2

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Centripetal acc. = ω²r = 4π²v²r
= 4 × (3.14)² × 120 × 30 = 23.7 ms^-2
60 100

[ ∵ ω = 2πv ]

Correct Option: C

Centripetal acc. = ω²r = 4π²v²r
= 4 × (3.14)² × 120 × 30 = 23.7 ms^-2
60 100

[ ∵ ω = 2πv ]

A particle moves in a circle of radius 5 cm with constant speed and time period 0.2 πs. The acceleration of the particle is

1. 15 m/s²
1. 25 m/s²
1. 36 m/s²
1. 5 m/s²

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Centripetal acceleration a_c = ω²r
= 2π ² r
T

= 2π ² × 5 × 10^-2 = 5 m / s²
0.2π

Correct Option: D

Centripetal acceleration a_c = ω²r
= 2π ² r
T

= 2π ² × 5 × 10^-2 = 5 m / s²
0.2π

The position vector of a particle R^{^→} as a function of time is given by :
R^{^→} = 4sin(2πt)î + 4cos(2πt)ĵ
Where R is in meter, t in seconds and î and ĵ denote unit vectors along x-and y-directions, respectively.
Which one of the following statements is wrong for the motion of particle?

1. Magnitude of acceleration vector is v² , where v is the velocity of particle
  R
1. Magnitude of the velocity of particle is 8 meter/second
1. path of the particle is a circle of radius 4 meter.
1. Acceleration vector is along R^{^→}

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Here,x = 4sin(2πt)...(i)
y = 4cos(2πt)...(ii)
Squaring and adding equation (i) and (ii)
x² + y² = 4² ⇒ R = 4
Motion of the particle is circular motion, accelerationvector is along – R^{^→} and its magnitude
= V²
R

Velocity of particle, V = ωR = (2π) (4) = 8π

Correct Option: B

Here,x = 4sin(2πt)...(i)
y = 4cos(2πt)...(ii)
Squaring and adding equation (i) and (ii)
x² + y² = 4² ⇒ R = 4
Motion of the particle is circular motion, accelerationvector is along – R^{^→} and its magnitude
= V²
R

Velocity of particle, V = ωR = (2π) (4) = 8π

A ship A is moving Westwards with a speed of 10 km h^–1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h^–1. The time after which the distance between them becomes shortest, is :

1. 5 h
1. 5 √2 h
1. 10 √2 h
1. 0 h

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V^{^→}_A = 10(-î)
V^{^→}_A = 10(ĵ)
V^{^→}_BA = 10ĵ + 10î = 10 √2 km / h
Distance OB = 100 cos 45° = 50 √2 km

Time taken to reach the shortest distance between A and B
= OB^{^→} = 50 √2 = 5h
V^{^→}_BA 10 √2

Correct Option: A

V^{^→}_A = 10(-î)
V^{^→}_A = 10(ĵ)
V^{^→}_BA = 10ĵ + 10î = 10 √2 km / h
Distance OB = 100 cos 45° = 50 √2 km

Time taken to reach the shortest distance between A and B
= OB^{^→} = 50 √2 = 5h
V^{^→}_BA 10 √2

A particle moves so that its position vector is given by r^{^→} = cos ωtx̂ + sin ωtŷ . Where ω is a constant. Which of the following is true?

1. Velocity and acceleration both are perpendicular to r^{^→}
1. Velocity and acceleration both are parallel to r^{^→}
1. Velocity is perpendicular to r^{^→} and acceleration is directed towards the origin
1. Velocity is perpendicular to r^{^→} and acceleration is directed away from the origin

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Given : Position vector r^{^→} = cos ωt x̂ + sin ωt ŷ
∴ Velocity , v^{^→} = – ωsin ωt x̂ + ωcos ωt ŷ
and acceleration , a^{^→} = –ω² cos ωt x̂ – ω²sin ωt ŷ = – ω² r^{^→}
r^{^→} . v^{^→} = 0 hence r^{^→} ⊥ v^{^→} and
a^{^→} is directed towards the origin.

Correct Option: C

Given : Position vector r^{^→} = cos ωt x̂ + sin ωt ŷ
∴ Velocity , v^{^→} = – ωsin ωt x̂ + ωcos ωt ŷ
and acceleration , a^{^→} = –ω² cos ωt x̂ – ω²sin ωt ŷ = – ω² r^{^→}
r^{^→} . v^{^→} = 0 hence r^{^→} ⊥ v^{^→} and
a^{^→} is directed towards the origin.