Problems

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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 = = $${100 \;{\:\rm m}\over 2\pi (1.5 \;{\:\rm m})}$$ = 10.6.

b)

$$\Delta$$$$\theta$$ = 2$$\pi$$n = $${s\over r}$$ = $${100 \;{\:\rm m} \over 1.5\;{\:\rm m}}$$ = 66.7  radians.

c)

Average angular velocity:

$$\overline{\omega}$$ = $${\Delta \theta \over \Delta t}$$ = $${66.7 \;{\:\rm radians}\over 30 \;{\:\rm s}}$$ = 2.22  rads/s.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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:

$$\Delta$$$$\theta$$ = $$\omega_{o}^{}$$$$\Delta$$t + $${1\over 2}$$$$\alpha$$($$\Delta$$t)².

Using $\omega_{o}^{}$ = 0 and solving for $\alpha$ gives:

$$\alpha$$ = $${\frac{2\Delta\theta}{(\Delta t)^{2}}}$$ = $${\frac{2(66.7)}{30^2}}$$ = 0.15 rads/s ².

b)

$$\omega$$ = $$\omega_{o}^{}$$ + $$\alpha$$$$\Delta$$t = 0 + (0.15)(30) = 4.5  rads/s.

c)

After one revolution, $\Delta$$\theta$ = 2$\pi$ . Using

$$\omega^{2}_{}$$ = 2$$\alpha$$$$\Delta$$$$\theta$$ + $$\omega_{o}^{2}$$

we get

$$\omega$$ = $$\sqrt{2 \alpha (2\pi)}$$ = 1.37  rads/s.

The tangential velocity and acceleration are:

v_t = r$$\omega$$ = (1.5  m )(1.37  rads/s ) = 2.06  m/s

and

a_t = r$$\alpha$$ = (1.5  m )(0.15 rads/s ²) = .225 m/s ².

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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)

  

Figure 7.4: Problem 7.3b)

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.

ma_c = T + mg

so that

$$\omega^{2}_{}$$ T =

= .1  kg x .8  m x (4  rads/s )² - .1  kg x 9.8  m/s ² = 0.3  N.

b)

  

Figure 7.5: Problem 7.3b)

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

ma_c = T - mg

so that

T = ma_c + mg = mr$$\omega^{2}_{}$$ + mg = 2.26  N.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

The same yo-yo as in the previous problem is spun in a horizontal circle, as shown in Figure 7.5.

  

Figure 7.6: Problem 7.4

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$$\vec{a}_{c}^{}$$ = $$\vec{T}$$ + $$\vec{W}$$.

Decompose forces into vertical and horizontal components.

Vertical:

0 = Tcos $$\theta$$ - mg.

This implies that

T = $${\frac{mg}{\cos\theta}}$$ = 1.13  N.

Horizontal:

$$\theta$$ ma_c =

$$\Rightarrow \perp \omega^{2}_{} \theta$$ =

This gives:

$$\omega$$ = $$\sqrt{T \sin\theta\over m r_\perp}$$

where r_$\perp$ = rsin $\theta$ . So:

$$\omega$$ = $$\sqrt{T \sin\theta\over m r\sin\theta}$$ = 3.76  rads/s.

The period is therefore:

T = $${2\pi\over \omega}$$ = 1.67  s.

Note: as $\theta$$\to$90 ^o , T$\to$$\infty$ in order to balance the gravitational force, so that the period $\to$ 0: i.e. you must spin it infinitally fast to get it to be horizontal. Try it!

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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:

$$\sum {Gm M_E\over r^2}$$ =

$$\omega^{2}_{} {Gm M_E\over r^2.}$$ =

where M_E is the mass of the Earth. Solve for $\omega$ :

$$\omega$$ = $$\sqrt{G M_E \over r^3.}$$

The period is:

T = $${2\pi\over \omega}$$ = .

The above expression is called Kepler's Law. Note: The period is independent of the mass of the satelite.

$\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

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 = $\infty$ with v_f = 0 . The final kinetic and potential energy are zero, whereas KE_i = ${1\over 2}$mv_E² and PE_i = - GmM_E/R_E . Thus

0 = $${1\over 2}$$mv_E² - $${GmM_E\over R_E}$$

The escape velocity is therefore:

v_E = $$\sqrt{2 GM_E \over R_E}$$.

Using M_E = 6 x 10²⁴  kg and R_E = 6.4 x 10⁶  m yields:

v_E = 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:

  v_E = $$\sqrt{2 GM \over R}$$.

• 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, R_c , given by:

  R_c = $${2GM\over c^2}$$.

  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.