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A bicycle wheel of radius r = 1.5 m starts from rest and rolls 100 m without slipping in 30 s. Calculate a) the number of revolutions the wheel makes, b) the number of radians through which it turns, c)The average angular velocity.
n = = = 10.6.
= 2n = = = 66.7 radians.
= = = 2.22 rads/s.
Assuming that the angular acceleration of the wheel given above was constant, calculate: a) The angular acceleration, b) the final angular velocity c) the tangential velocity and tangential acceleration of a point on the rim after one revolution.
= t + (t)2.Using = 0 and solving for gives:
= = = 0.15 rads/s 2.
= + t = 0 + (0.15)(30) = 4.5 rads/s.
= 2 +we get
= = 1.37 rads/s.The tangential velocity and acceleration are:
vt = r = (1.5 m )(1.37 rads/s ) = 2.06 m/sand
at = r = (1.5 m )(0.15 rads/s 2) = .225 m/s 2.
A yo-yo of mass 100 g is spun in a vertical circle at the end of a 0.8 m string at a constant angular velocity of 4 rads/s. Calculate a) the tension in the string at the top of its trajectory, b) the tension in the string at the bottom of the trajectory.
mac = T + mgso that
|T||=||mr - mg|
|=||.1 kg x .8 m x (4 rads/s )2 - .1 kg x 9.8 m/s 2 = 0.3 N.|
mac = T - mgso that
T = mac + mg = mr + mg = 2.26 N.
The same yo-yo as in the previous problem is spun in a horizontal circle, as shown in Figure 7.5.
If the string makes an angle of 30 o to the vertical, what is the period of rotation?
Apply Newton's Laws of Motion:
m = + .Decompose forces into vertical and horizontal components.
0 = Tcos - mg.This implies that
T = = 1.13 N.
=where r = rsin . So:
= = 3.76 rads/s.The period is therefore:
T = = 1.67 s.
Note: as 90 o , T in order to balance the gravitational force, so that the period 0: i.e. you must spin it infinitally fast to get it to be horizontal. Try it!
What is the relationship between the radius of orbit of a satelite (mass m ) and its period?
The centripetal force is provided by the gravitational pull of the Earth, which is the only force acting on the satellite Newton's Second Law is therefore:
|Fc = mac||=|
=The period is:
T = = .The above expression is called Kepler's Law. Note: The period is independent of the mass of the satelite.
The escape velocity of any object is the speed it must achieve to escape the gravitational pull of the Earth. Calculate the escape velocity for an object of mass m .
We can use conservation of mechanical energy (neglecting air resistance):
KEf + PEf = KEi + PEi.Suppose the object starts at the earth's surface with speed vE and reaches r = with vf = 0 . The final kinetic and potential energy are zero, whereas KEi = mvE2 and PEi = - GmME/RE . Thus
0 = mvE2 -The escape velocity is therefore:
vE = .Using ME = 6 x 1024 kg and RE = 6.4 x 106 m yields:
vE = 11 km/sNote:
vE = .
Rc = .When an object collapses below its critical radius, it becomes a black hole, from which nothing, not even light, can escape. The critical radius for the sun is about 3 km, while the critical radius for the earth is about 5 cm.