Find
i) Time of pure rolling
ii) velocity of plank
iii) velocity of disc
iv) angular velocity of disc
v) work done by friction
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Sliding and Rolling disc m on top of a rough plank M resting on a smooth surface...
Mass of Disc = m = 1kg. Mass of Plank = M = 2kg. Radius = R = 1m.
Coefficient of kinetic Friction = μ = 0.50. This is a high value really.
Friction force = f, acts in the backward direction to slow down the disc. Friction f acts on the plank in the forward direction.
Initial angular velocity = 0. Initial speed of disc = u = 10m/s.
Angular velocity at time t = ω. Angular acceleration = α.
Linear acceleration of the disc = a.
Linear acceleration of the plank = a2. Initial speed of plank = 0.
Friction between the plank and floor = 0. Smooth floor.
Plank is a moving non inertial frame of reference. So a pseudo force (linear) = - m *a2 is to be added to the disc m.
For the plank: f = M * a2 => a2 = f/M --- (1)
Torque on the disc about its center O = T = I α = f R
So m R²/2 * α = f R => α = 2f/(m R) --- (2)
For the disc, in the frame of reference of the accelerating plank:
Total force = - f - m * a2 = m * a
a = - f [ 1/m + 1/M] = - f (m+M)/(mM) --- (3)
v = u + a t = u - f (m+M) t /(mM) ..... v decreases ...
ω = ω₀ + α t = 0 + 2 f t/(m R) --- (4) .... ω increases ..
For Rolling to start, v = R ω --- (5)
Solving equations (4) and (5), we get the time t to start rolling.
t = m M u / (m + 2M) f --- (6)
Substituting this value in (4) we get:
v_r = 2 M u /(m+3M). ω_r = 2 M u /[ (m+3M) R ] --- (7)
velocity of the plank = v2 = a2 * t = f t /M = m u/(m+3M) --- (8)
v_r is relative to the plank. So absolute velocity v of the disc m is:
v = v_r + v2 = (m+2M) u /(m+3M) --- (9)
Initial KE = 1/2 m u²
Final KE = 1/2 m v² + 1/2 I ω_r² + 1/2 M v2²
= 1/2 m² u² /(m+3M)² * [ (m+2M)² + 2 m² + m M ]
= 1/2 m² u² /(m+3M)² * [m²+ 6M² + 5 m M ]
Work done by friction = Loss in KE
W_f = 1/2 m u² / (m+3M)² * [ (m+3M)² - m² - 6M² - 5 m M ]
= 1/2 m u² * [ M /(m + 3M) ]
Mass of Disc = m = 1kg. Mass of Plank = M = 2kg. Radius = R = 1m.
Coefficient of kinetic Friction = μ = 0.50. This is a high value really.
Friction force = f, acts in the backward direction to slow down the disc. Friction f acts on the plank in the forward direction.
Initial angular velocity = 0. Initial speed of disc = u = 10m/s.
Angular velocity at time t = ω. Angular acceleration = α.
Linear acceleration of the disc = a.
Linear acceleration of the plank = a2. Initial speed of plank = 0.
Friction between the plank and floor = 0. Smooth floor.
Plank is a moving non inertial frame of reference. So a pseudo force (linear) = - m *a2 is to be added to the disc m.
For the plank: f = M * a2 => a2 = f/M --- (1)
Torque on the disc about its center O = T = I α = f R
So m R²/2 * α = f R => α = 2f/(m R) --- (2)
For the disc, in the frame of reference of the accelerating plank:
Total force = - f - m * a2 = m * a
a = - f [ 1/m + 1/M] = - f (m+M)/(mM) --- (3)
v = u + a t = u - f (m+M) t /(mM) ..... v decreases ...
ω = ω₀ + α t = 0 + 2 f t/(m R) --- (4) .... ω increases ..
For Rolling to start, v = R ω --- (5)
Solving equations (4) and (5), we get the time t to start rolling.
t = m M u / (m + 2M) f --- (6)
Substituting this value in (4) we get:
v_r = 2 M u /(m+3M). ω_r = 2 M u /[ (m+3M) R ] --- (7)
velocity of the plank = v2 = a2 * t = f t /M = m u/(m+3M) --- (8)
v_r is relative to the plank. So absolute velocity v of the disc m is:
v = v_r + v2 = (m+2M) u /(m+3M) --- (9)
Initial KE = 1/2 m u²
Final KE = 1/2 m v² + 1/2 I ω_r² + 1/2 M v2²
= 1/2 m² u² /(m+3M)² * [ (m+2M)² + 2 m² + m M ]
= 1/2 m² u² /(m+3M)² * [m²+ 6M² + 5 m M ]
Work done by friction = Loss in KE
W_f = 1/2 m u² / (m+3M)² * [ (m+3M)² - m² - 6M² - 5 m M ]
= 1/2 m u² * [ M /(m + 3M) ]
kvnmurty:
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