Question

A small sphere of mass m suspended by a thread is first taken aside so that the thred forms the right angle with the virtical and then released. Find:
(a)the total accleration of the sphere and the thread tension as a function of θ, the angle of defliction of the thred from the virtical.

Answer 1

Answer on Question #45307, Physics, Mechanics | Kinematics | Dynamics

A small sphere of mass $m$ suspended by a thread is first taken aside so that the thread forms the right angle with the vertical and then released. Find:

(a) the total acceleration of the sphere and the thread tension as a function of $\theta$ , the angle of deflection of the thread from the vertical.

Solution:

Mechanical energy $E$ is the sum of the potential and kinetic energies of an object.

$$E = U + K$$

The total mechanical energy in any isolated system of objects remains constant if the objects interact only through conservative forces:

$$U _ {f} + K _ {f} = U _ {i} + K _ {i}$$ $$m g l = \frac {m v ^ {2}}{2} + m g l (1 - \cos \theta)$$

where $l$ is length of the thread.

$$g l = \frac {v ^ {2}}{2} + g l (1 - \cos \theta)$$ $$v ^ {2} = 2 g l \cos \theta$$

The normal acceleration is

$$a _ {n} = \frac {v ^ {2}}{l} = 2 g \cos \theta$$

The linear displacement from equilibrium is $s$ , the length of the arc. Also on figure shown are the forces on the ball, which result in a net force of $F = mgsin\theta$ toward the equilibrium position—that is, a restoring force.

The tangential acceleration is

$$a _ {t} = \frac {F}{m} = g \sin \theta$$

The total acceleration is

$$a = \sqrt {a _ {n} ^ {2} + a _ {t} ^ {2}} = g \sqrt {4 \cos^ {2} \theta + \sin^ {2} \theta} = g \sqrt {3 \cos^ {2} \theta + \cos^ {2} \theta + \sin^ {2} \theta} = g \sqrt {3 \cos^ {2} \theta + 1}$$

The force equation is

$$m g \cos \theta = m a _ {n}$$

Thus,

$$= m a _ {n} + m g \cos \theta = 2 m g \cos \theta + m g \cos \theta = 3 m g \cos \theta$$

Answer: $a = g\sqrt{3\cos^2\theta + 1}$ , $= 3mg\cos \theta$ .

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