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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.
Solution:
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.
Solution:
= 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:v_{t} = r = (1.5 m )(1.37 rads/s ) = 2.06 m/s
anda_{t} = 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.
Solution:
ma_{c} = T + mg
so thatT | = | mr - mg | |
= | .1 kg x .8 m x (4 rads/s )^{2} - .1 kg x 9.8 m/s ^{2} = 0.3 N. |
ma_{c} = T - mg
so thatT = ma_{c} + 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?
Solution:
Apply Newton's Laws of Motion:
m = + .
Decompose forces into vertical and horizontal components.Vertical:
0 = Tcos - mg.
This implies thatT = = 1.13 N.
Horizontal:
ma_{c} | = | Tsin | |
mr_{} | = | Tsin . |
=
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?
Solution:
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:
F_{c} = ma_{c} | = | ||
mr | = |
=
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 .
Solution:
We can use conservation of mechanical energy (neglecting air
resistance):
KE_{f} + PE_{f} = KE_{i} + PE_{i}.
Suppose the object starts at the earth's surface with speed v_{E} and reaches r = with v_{f} = 0 . The final kinetic and potential energy are zero, whereas KE_{i} = mv_{E}^{2} and PE_{i} = - GmM_{E}/R_{E} . Thus0 = mv_{E}^{2} -
The escape velocity is therefore:v_{E} = .
Using M_{E} = 6 x 10^{24} kg and R_{E} = 6.4 x 10^{6} m yields:v_{E} = 11 km/s
Note:v_{E} = .
R_{c} = .
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.www-admin@theory.uwinnipeg.ca