Question

A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 2° and time period 2 s. Find (a). the radius of the circular wire, (b). the acceleration of this particle as it goes through its mean position and (c). the acceleration of this particle when it is at an extreme position. Take g = π² m/s².

Answer 1

Explanation ⇒

(a). Given case is of Physical pendulum.

∴ $T = 2\pi \sqrt{\frac{i}{mgl} }$,

where i is the moment of inertia of wire about the point of suspension.

Since, r = l in this case.

Thus, Moment of inertia = i = 2mr²

∴ $T = 2\pi \sqrt{\frac{2mr^{2} }{mgr} }\\T = 2\pi \sqrt{\frac{2r}{g} }$

Since, T = 2 seconds.

Thus, $4 = 4\pi^{2} \frac{2r}{g} \\2r\pi^2 = g$

Now, g =  π² m/s²

∴ 2r = g/ π²

∴ 2r = 1

∴ r = 0.5 m.

Hence, the radius of the circular wire is 0.5 m.

______________________________

(b). When the particle goes through the mean position, then in that case  there is no torque on the wire since the centre of mass and the point of suspension are in the same vertical line.

As a result there will be no acceleration due to the S.H.M.

But we should not forget that  the farthest point is moving in circle with radius 2r having its center at the point of suspension, thus it will experiences an centripetal acceleration.

∴ Using the formula of Centripetal acceleration

$a_c = \frac{v^2}{r}$

∴ a = 0.11²/(2 × 0.5)

∴ a = 121/10000

∴ a = 1.21/100

∴ a = 0.0121 m/s².

Hence, the acceleration of the particle through the mean position is 0.0121 m/s².

______________________________

(c). As we know that at the extreme position velocity of the particle is zero, thus, no centripetal acceleration will be there.

But there is a torque on the wire, since, Center of mass and point of suspension will now not be parallel to each other.

Using the formula of the torque,

τ = mgrsinθ

∴ iα = mgrsinθ

∴ α = mgrsinθ/i

∴ α = mgrsinθ/2mr²

∴  α = gsinθ/2r

Now,  α = = a/r

∴ a/2r = gsinθ/2r

∴ a = gsinθ

∴ a = π²Sin2°.

∴ a = (3.14)² × 0.034

∴ a = 0.34 m/s².

Hence, the acceleration through the extreme position is 0./34 m/s².

Hope it helps.

Answer 2

Explanation ⇒

(a). Given case is of Physical pendulum.

∴ T = 2\pi \sqrt{\frac{i}{mgl} }T=2π

mgl

i

,

where i is the moment of inertia of wire about the point of suspension.

Since, r = l in this case.

Thus, Moment of inertia = i = 2mr²

∴ \begin{lgathered}T = 2\pi \sqrt{\frac{2mr^{2} }{mgr} }\\T = 2\pi \sqrt{\frac{2r}{g} }\end{lgathered}

T=2π

mgr

2mr

2

T=2π

g

2r

Since, T = 2 seconds.

Thus, \begin{lgathered}4 = 4\pi^{2} \frac{2r}{g} \\2r\pi^2 = g\end{lgathered}

4=4π

2

g

2r

2rπ

2

=g

Now, g = π² m/s²

∴ 2r = g/ π²

∴ 2r = 1

∴ r = 0.5 m.

Hence, the radius of the circular wire is 0.5 m.

______________________________

(b). When the particle goes through the mean position, then in that case there is no torque on the wire since the centre of mass and the point of suspension are in the same vertical line.

As a result there will be no acceleration due to the S.H.M.

But we should not forget that the farthest point is moving in circle with radius 2r having its center at the point of suspension, thus it will experiences an centripetal acceleration.

∴ Using the formula of Centripetal acceleration

a_c = \frac{v^2}{r}a

c

=

r

v

2

∴ a = 0.11²/(2 × 0.5)

∴ a = 121/10000

∴ a = 1.21/100

∴ a = 0.0121 m/s².

Hence, the acceleration of the particle through the mean position is 0.0121 m/s².

______________________________

(c). As we know that at the extreme position velocity of the particle is zero, thus, no centripetal acceleration will be there.

But there is a torque on the wire, since, Center of mass and point of suspension will now not be parallel to each other.

Using the formula of the torque,

τ = mgrsinθ

∴ iα = mgrsinθ

∴ α = mgrsinθ/i

∴ α = mgrsinθ/2mr²

∴ α = gsinθ/2r

Now, α = = a/r

∴ a/2r = gsinθ/2r

∴ a = gsinθ

∴ a = π²Sin2°.

∴ a = (3.14)² × 0.034

∴ a = 0.34 m/s².

Hence, the acceleration through the extreme position is 0./34 m/s².