An automobile travelling at 80 km hr-1 has,tyres of radius 80 cm. on applying breakes, the car is brought to a stop in 30 complete turns of the tyres. What is the magnitude of the angular acceleration of the wheels ? How far does the car move while the breakes are applied ?
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u = 80 kmph = 80 * 5/18 = 200/9 m/s
radius r = 80 cm = 0.80 m
Distance traveled = 30 rotations = 30 * 2 π r = 48 π m
Initial angular velocity = ω0 = u/r = 200/9 * 1/0.80 = 200/7.2 = 250/9 rad/s
final angular velocity = ω1 = 0 stopped
Ф = total angle turned = 30 turns * 2π rad/turn = 60π rad
α = angular acceleration
ω1² = ω0² + 2 α Ф
0 = (250/9)² + 2 α * 60 π
α = - 2.046 rad/sec²
radius r = 80 cm = 0.80 m
Distance traveled = 30 rotations = 30 * 2 π r = 48 π m
Initial angular velocity = ω0 = u/r = 200/9 * 1/0.80 = 200/7.2 = 250/9 rad/s
final angular velocity = ω1 = 0 stopped
Ф = total angle turned = 30 turns * 2π rad/turn = 60π rad
α = angular acceleration
ω1² = ω0² + 2 α Ф
0 = (250/9)² + 2 α * 60 π
α = - 2.046 rad/sec²
kvnmurty:
we could also find linear acceleration a and divide by r. a = u^2/2s = (80*5/18)^2 / [2 * 30 * 2 pi r ] ... alpha = a / r
kvnmurthy friend main ne sabhi question ek saath post kiya pad lo aap . makn ek ek questions ko post kar diya hai par aap ko wo questions understand nahi ho ga
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