Find the time period of small oscillations of the following systems. (a) A meterstick suspended through the 20 cm mark. (b) A ring of mass m and radius r suspended through a point on its periphery. (c) A uniform square plate of edge a suspended through a corner. (d) A uniform disc of mass m and radius r suspended through a point r/2 away from the center.
Answers
Answer ⇒ (a). T = 1.51 s , (b). T = 2π√(2r/g) , (c). T = 2π√{√8a/3g} and T = 2π√(3r/2g)
Explanation ⇒ (a) When the meter stick is suspended through 20 cm mark, then its motion can't be linear. Thus, its motion is angular and that too angular S.H.M.
Thus, it is a physical pendulum.
Thus, Using the formula,
, where i is the moment of inertia of the rod about its end.
now, i about the end of centre = mL²/12.
Using parallel axis theorem, Moment of inertia about its end = mL²12 + ml²,
l = 100 - (50 + 20) = 30 cm. [It is suspended on 20 cm mark. At 50 cm, COM acts.]
∴ M.I. about the horizontal axis through the point of support =
= mL²/12 + ml² [By Parallel axis theorem.]
= m(1/12+0.30²) {Since L = 1}
= m(0.083+0.09)
= 0.173m
Thus, Using the formula,
T = 2π√{0.173m/(m × 9.8 × 0.30)}
=2π√(0.173/9.8*0.3)
= 2π√0.058
= 1.51 s
This is the required time period of the physical pendulum.
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(b). We know that the centre of mass of the ring of the ring is at the center of the ring. Also, as per as the question, the point of suspension is at the periphery of the ring.
Hence l = r.
Moment of inertia of the ring about a horizontal axis through the center = mr²
Quantity to be added in moment of inertia = ml² = mr² [Since, l = r.]
Applying parallel axis theorem,
M.I. of the ring about a horizontal axis through the support =mr²+mr²
= 2mr²
∴ The time period = 2π√(i/mgl)
= 2π√(2mr²/mgr)
= 2π√(2r/g)
Hence, the time period will be 2π√(2r/g) .
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(c). For the square plate hanged through a corner, l = a/√2
Thus, again applying the parallel axis theorem,
Moment of inertia about a horizontal axis through the point of suspension
= ma²/6 + ml²
= ma²/6 + m(a/√2)²
= m{a²/6+a²/2}
= 2ma²/3
Hence the time period = 2π√{(2ma²/3)/(mga/√2)}
∴ T = 2π√{2√2a/3g}
T = 2π√{√8a/3g}
Hence, the time period in this case is T = 2π√{√8a/3g}.
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(d). In the case of the uniform disc suspended through a point at a distance r/2 away from the center, l = r/2.
Thus, applying Parallel axis theorem again,
∴ Moment of inertia about the horizontal axis through the point of suspension = 1/2 × mr²+ ml²
= 1/2 × mr² + m(r/2)²
=1/2 × mr²+ 1/4 × mr²
=3/4 × mr²
∴ The time period =2π√{(3/4mr²)/(mgr/2)}
= 2π√(3r/2g)
Hence, the time period in this case is 2π√(3r/2g)
Hope it helps.
Explanation:
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