A 2000-kg car accelerates uniformly from rest to 120 km/hr in 10 seconds, then m
ID: 2061359 • Letter: A
Question
A 2000-kg car accelerates uniformly from rest to 120 km/hr in 10 seconds, thenmaintains constant speed. This happens on the equator and the car is moving in the opposite
direction of the roads velocity vector. (The car is moving against the rotation of the earth) The
radius of the earth is 6.37 x 106 m and assume the earth’s density is uniformly distributed as a
solid sphere of mass 5.98 x 1024 kg. Before this acceleration the earth’s angular velocity is 1 rev
per day.
A.) Does this increase, decrease, or leave the period of a day unchanged?
For parts B through D only algebraic equations in terms of the given information should be
presented, no numbers.
B.) What is the linear acceleration of the car?
C.) What is the net torque, due to the car, applied to the earth during the acceleration?
D.) What is the final angular velocity of the earth.
E.) What is the ratio of the amount of time for a day before this acceleration versus after?
(ignore the 10 seconds the car accelerated.) This requires many decimal points so keeping the
equations algebraic until the very end avoids round off error during the intermittent steps
thank you, will rate
Explanation / Answer
A) this decreases the period of the day
B) acceleration,a=(v-u)/t = (final velocity-initial velocity)/time taken=(120*5/18)/10=3.33m/s^2
C) force applied by earth on car = ma = 2000*3.333333=6666.66N
therfore force applied by car on earth is also the same = 6666.66N
Torque = force*distance = 6666.66*6.37*10^6 = 42466.62*10^6N-m
D)angular velocity=1rev/day = 2rad/86400seconds = 7.27*10^-5rad/s
E)moment of inertia of earth = (2/5)MR^2
change in angular welocity = t/I = maR(10)/[(2/5)MR^2] = 25ma/MR
1 = 2/d1 d1=time take for day before accelration=86400seconds
2 = 2/d2 d2=time take for day after accelration
2-1 = 2(1/d2 - 1/d1) = 25ma/MR
multiply with d1
2(d1/d2) = 2 + 25ma(d1)/MR
d1/d2 = 1+ 25ma(d1)/2MR
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