# Problems

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:

a)
If there is no slipping, the arc length through which a point of the rim moves is equal to the distance travelled, so that the number of revolutions is:

n = = = 10.6.

b)

= 2n = = = 66.7  radians.

c)
Average angular velocity:

= = = 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:

a)
For constant angular acceleration:

= t + (t)2.

Using = 0 and solving for gives:

= = = 0.15 rads/s 2.

b)

= + t = 0 + (0.15)(30) = 4.5  rads/s.

c)
After one revolution, = 2 . Using

= 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/s

and

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.

Solution:

a)

At the top of its trajectory, the centripetal force is provided by the sum of the tension in the string and the weight, see Figure 7.4.

mac = T + mg

so 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.
b)

At the bottom of the trajectory, the centripetal force is provided by the difference between the tension and the weight (see Figure 7.5).

mac = T - mg

so 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?

Solution:
Apply Newton's Laws of Motion:

m = + .

Decompose forces into vertical and horizontal components.

Vertical:

0 = Tcos - mg.

This implies that

T = = 1.13  N.

Horizontal:

 mac = Tsin mr = Tsin .
This gives:

=

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:

 Fc = mac = mr =
where ME is the mass of the Earth. Solve for :

=

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):

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/s

Note:
• The escape velocity is independent of the mass of the object.
• This excape velocity of any object a distance R from the center of any planet or star of mass M , can be calculated in exactly the same way to give:

vE = .

• If the radius of the planet or star of fixed mass M gets very small, the escape velocity of an object at its surface can exceed the speed of light. This will happen if the radius goes below a critical radius, Rc , given by:

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.

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