Question
A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in what will be the trajectory of the particle if the string is cut at:
- Point B?
- Point C?P
- oint X?
✓
Answer
Key concept: According to the situation shown above that a bob of mass m is whirled into a vertical circle, the required centripetal force is obtained from the net force towards center at any point of time in the string. Tension in the string is variable and it is always towards center. But the gravitational force on the bob is always towards center. The speed of the body will be different at different points. So the equations of dynamical equilibrium $(F_c = ma_c, F_t = ma_t)$ must be satisfied at all the points. Let when the string makes an angle 9 with vertical, the speed of mass is v. Apply Newton’s law perpendicular to the string:$\text{Mg}\sin=\text{ma}$
$\Rightarrow\text{a}=\text{g}\sin$
The above equation gives tangential acceleration as a function of angle . At lowest point = 0° and at highest point = 180°. So at both points sin 9 = 0. Hence a, = 0 at both points L and H. At point M, = 90°, then $a_1 = g$. It is the maximum value of $a_t$ Apply Newton’s law along the string:$\text{T}-\text{mg}\cos=\text{ma}$
Or $\text{T}=\text{mg}\cos+\text{m}\frac{\text{v}^2}{\text{r}}\ ...(\text{i})$ As the body goes up, its velocity will go on decreasing and angle $\theta$ will go on increasing. Maximum speed of the body will be at lowest point L and minimum at highest point H. Then from above relation we can find that tension will be maximum at lowest point and minimum at highest point. Tension at lowest point $(\theta=0^\circ,\text{v}=\text{v}_\text{L})$$\text{T}_\text{L}=\text{mg}+\text{m}\frac{\text{v}^2_\text{L}}{\text{r}}\ ...(\text{ii})$
Tension at highest point $(\theta=180^\circ,\text{v}=\text{v}_\text{H})$$\text{T}_\text{H}=-\text{mg}+\text{m}\frac{\text{v}^2_\text{H}}{\text{r}}\ ...(\text{iii})$
When string is cut, tension in string becomes zero and centripetal force is not provided. Hence, bob tends to move in along the direction of its velocity.
$\Rightarrow\text{a}=\text{g}\sin$
The above equation gives tangential acceleration as a function of angle . At lowest point = 0° and at highest point = 180°. So at both points sin 9 = 0. Hence a, = 0 at both points L and H. At point M, = 90°, then $a_1 = g$. It is the maximum value of $a_t$ Apply Newton’s law along the string:$\text{T}-\text{mg}\cos=\text{ma}$
Or $\text{T}=\text{mg}\cos+\text{m}\frac{\text{v}^2}{\text{r}}\ ...(\text{i})$ As the body goes up, its velocity will go on decreasing and angle $\theta$ will go on increasing. Maximum speed of the body will be at lowest point L and minimum at highest point H. Then from above relation we can find that tension will be maximum at lowest point and minimum at highest point. Tension at lowest point $(\theta=0^\circ,\text{v}=\text{v}_\text{L})$$\text{T}_\text{L}=\text{mg}+\text{m}\frac{\text{v}^2_\text{L}}{\text{r}}\ ...(\text{ii})$
Tension at highest point $(\theta=180^\circ,\text{v}=\text{v}_\text{H})$$\text{T}_\text{H}=-\text{mg}+\text{m}\frac{\text{v}^2_\text{H}}{\text{r}}\ ...(\text{iii})$
When string is cut, tension in string becomes zero and centripetal force is not provided. Hence, bob tends to move in along the direction of its velocity.
- If the string is cut at any point, then velocity of body of mass m is along the tangent to the circle. Tangent at point B is vertically downward so the trajectory of the particle is the straight line.
- Tangent at point C is horizontally towards right. So the trajectory of the particle is the parabola.
- Tangent at point X makes some angle with the horizontal. Again bob will follow a parabolic path with vertex higher than C.
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