A 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of 0.10 N - m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its time period of the oscillation will be:

Question

A 25 kg uniform solid sphere with a 20 cm radius is suspended by a vertical wire such that the point of suspension is vertically above the centre of the sphere. A torque of 0.10 N - m is required to rotate the sphere through an angle of 1.0 rad and then maintain the orientation. If the sphere is then released, its time period of the oscillation will be: (A) π second (B) √2 π second (C) 2 π second (D) 4 π second

Tardigrade Admin · Accepted Answer

τ=-k θ 0.1=-k(1.0), where k is torsional constant of the wire. k =(1/10) T =2 π √(I/k) =2 π √((2/5) × 25 ×(.2)2/1 / 10) =2 π √10 × .2 × .2 × 10=4 π sec

$π$ second

$2 π$ second

$2 π$ second

$4 π$ second

$τ = - k θ$
$0.1 = - k (1.0)$ , where $k$ is torsional constant of the wire.
$k = \frac{1}{10}$
$T = 2 π \frac{I}{k}$
$= 2 π \frac{\frac{2}{5} \times 25 \times ( .2 ) ^{2}}{1/10}$
$= 2 π 10 \times .2 \times .2 \times 10 = 4 π sec$

π second

2​π second

2π second

4π second

τ=−kθ 0.1=−k(1.0), where k is torsional constant of the wire. k=101​ T=2πkI​​ =2π1/1052​×25×(.2)2​​ =2π10×.2×.2×10​=4πsec

$π$ second

$2 π$ second

$2 π$ second

$4 π$ second

$τ = - k θ$
$0.1 = - k (1.0)$ , where $k$ is torsional constant of the wire.
$k = \frac{1}{10}$
$T = 2 π \frac{I}{k}$
$= 2 π \frac{\frac{2}{5} \times 25 \times ( .2 ) ^{2}}{1/10}$
$= 2 π 10 \times .2 \times .2 \times 10 = 4 π sec$