Question

Ques no. 2 &4 on
http://i66.tinypic.com/mli79k.jpg

Answer 1

Answer on Question #56588-Physics-Mechanics-Relativity

5-2 A ball of radius $R = 10$ cm rolls without slipping down an inclined plane so that its centre moves with constant acceleration $2.5 \, \text{cm/s}^2$ . After a time of 2 sec from the beginning of its motion, its position is as shown in figure-5.113. Find

(a) the velocities of point $A$ , $B$ and $O$ .

(b) the acceleration of these points.

Ans. [10 cm/s, 7.1 cm/s, 0, 5.6 cm/s², 2.5 cm/s², 2.5 cm/s²]

Figure 5.113

Solution

(a) No slipping:

$$\omega = \frac {v _ {t r}}{R} = \frac {a t}{R}.$$

For the points on the circumstance $v_{tr} = v_{rot} = \omega R$ . For the center $v_{rot} = \omega \cdot 0 = 0$ .

$$v _ {A} = v _ {t r} + v _ {r o t} = 2 v _ {t r} = 2 a t = 2 \cdot 2. 5 \cdot 2 = 1 0 \frac {c m}{s}.$$ $$v _ {B} = \sqrt {v _ {t r} ^ {2} + v _ {r o t} ^ {2}} = \sqrt {2} v _ {t r} = \sqrt {2} a t = \sqrt {2} \cdot 2. 5 \cdot 2 = 7. 1 \frac {c m}{s}.$$ $$v _ {O} = v _ {t r} - v _ {r o t} = 0.$$

(b) The acceleration is

$$\ddot {a} = \overline {{a _ {C}}} + \overline {{a _ {\tau}}} + \overline {{a _ {n}}}$$ $$a _ {C} = a.$$ $$a _ {n} = \frac {v _ {t r} ^ {2}}{R} = \frac {(2 . 5 \cdot 2) ^ {2}}{1 0} = 2. 5 \frac {c m}{s ^ {2}} = a.$$ $$a _ {\tau} = \frac {d v}{d t} = \frac {v _ {t r}}{t} = \frac {a t}{t} = a.$$

For A:

$$\overline {{a _ {C}}} = \overline {{a _ {\tau}}}.$$ $$a _ {A} = \sqrt {a ^ {2} + (2 a) ^ {2}} = \sqrt {5} a = \sqrt {5} \cdot 2. 5 = 5. 6 \frac {c m}{s ^ {2}}.$$

For B:

$$\overline {{a _ {C}}} + \overline {{a _ {n}}} = 0$$ $$a _ {B} = a _ {\tau} = a = 2. 5 \frac {c m}{s ^ {2}}.$$

For C:

$$\overline {{a _ {C}}} + \overline {{a _ {\tau}}} = 0$$ $$a _ {B} = a _ {n} = a = 2.5 \frac {cm}{s ^ {2}}.$$

5-4 A conical pendulum is formed by a thin rod of length $l$ and mass $m$, hinged at the upper end, rotates uniformly about a vertical axis passing through its upper end, with angular velocity $\omega$. Find the angle $\theta$ between the rod and the vertical.

$$\text{Ans. } [\theta = \cos^{-1} \left(\frac{3g}{2\omega^{2}l}\right)]$$

Solution

Looking at the rotating reference frame of the rod, there is a centrifugal force and gravity. Their torque must sum to zero because the rod is in equilibrium in that reference frame. For each bit of mass $dm$, we have

$$d \tau = (x \cos \theta) (\omega^{2} x \sin \theta) d m.$$

plugging in $dm = \left(\frac{m}{l}\right) dx$ and simplify, I get

$$d \tau = \frac{m \omega^{2} x^{2}}{l} \sin \theta \cos \theta dx.$$ $$\tau = \int_{0}^{l} \frac{m \omega^{2} x^{2}}{l} \sin \theta \cos \theta dx = \frac{1}{3} m \omega^{2} l^{2} \sin \theta \cos \theta.$$

The torque provided by gravity is simply $mg\left(\frac{l}{2}\right) \sin \theta$. Equating with the centrifugal torque, I get

$$\frac{1}{3} m \omega^{2} l^{2} \sin \theta \cos \theta = m g \left(\frac{l}{2}\right) \sin \theta$$ $$\cos \theta = \frac{3g}{2 \omega^{2} l}$$ $$\theta = \cos^{-1} \left[ \frac{3g}{2 \omega^{2} l} \right]$$

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