1) A body is moving along a horizontal circular path with constant speed v. What is the change in its velocity when it describes an angle?
{answer should come 2v sin}
2) Particle A of mass M is revolving along a circle of radius R. Particle B of mass m is revolving in another circle of radius r. If they take the same time to complete one revolution, then the ratio of their angular velocities is
{answer should come 1}
3) Two particles of masses M an m revolve in circular orbits of radii R and r. If the time periods of both the particles are equal, then the ratio of their linear speed is
{answer should come }
4) The angular velocity of a wheel increases from 600 revolutions/minute to 2400 revolutions/minute in 10 seconds. The number of revolutions made during this time interval is
{answer should come 250 revolutions}
5) A particle moves along a circle of radius 10 cm. If its linear speed changes from 4 m/s to
5 m/s in 1 second, then its angular acceleration will be
{answer should come 10}
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1. refer to diagram.
Change in velocity ΔV = V₂ - V₁ vector subtraction.
V1 and V2 are of same magnitude V but are inclined at angle Ф. The triangle formed by the three vectors is an Isosceles triangle. Draw a perpendicular bisector on to the vector V2 - V1. The angle is bisected.
Now magnitude of V2 - V1 = 2 * sin Ф/2 * hypotenuse = 2 V sin Ф/2
3. Let the time period of revolution of both masses be T.
In time T, m travels a distance = 2 π r and M moves a distance 2 π R.
So linear speed of m = 2π r / T linear speed of M = 2 π R / T
Ratio = R/r
2. Let the time period of revolution of both masses be T. In time T mass m covers an angle of 2 π radians and M also covers an angle of 2π rad.
angular speed = ω = 2π/T = same for both
4. Revolutions made = angle Ф rotated through in 10 seconds / 2π
Ф = time duration * average angular speed = 10 * (ω₂ + ω₁) / 2
= 10 * [ (2400 + 600)/2 ] * (2π/60) rad
= 500 π rad
n = revolutions = 500π/2π = 250
5. linear speed = v = r ω
angular speed ω = v / r
angular velocities ω2 = 5 /0.1 = 50 rad/sec ω1=4/0.1 = 40 rad/sec
angular acceleration = α = ω2 - ω1 / time = (50-40) / 1sec = 10 rad/sec²
Change in velocity ΔV = V₂ - V₁ vector subtraction.
V1 and V2 are of same magnitude V but are inclined at angle Ф. The triangle formed by the three vectors is an Isosceles triangle. Draw a perpendicular bisector on to the vector V2 - V1. The angle is bisected.
Now magnitude of V2 - V1 = 2 * sin Ф/2 * hypotenuse = 2 V sin Ф/2
3. Let the time period of revolution of both masses be T.
In time T, m travels a distance = 2 π r and M moves a distance 2 π R.
So linear speed of m = 2π r / T linear speed of M = 2 π R / T
Ratio = R/r
2. Let the time period of revolution of both masses be T. In time T mass m covers an angle of 2 π radians and M also covers an angle of 2π rad.
angular speed = ω = 2π/T = same for both
4. Revolutions made = angle Ф rotated through in 10 seconds / 2π
Ф = time duration * average angular speed = 10 * (ω₂ + ω₁) / 2
= 10 * [ (2400 + 600)/2 ] * (2π/60) rad
= 500 π rad
n = revolutions = 500π/2π = 250
5. linear speed = v = r ω
angular speed ω = v / r
angular velocities ω2 = 5 /0.1 = 50 rad/sec ω1=4/0.1 = 40 rad/sec
angular acceleration = α = ω2 - ω1 / time = (50-40) / 1sec = 10 rad/sec²
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shaggyy:
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