A solid cylinder of mass m and radius R is kept in equilibrium on horizontal rough surface. Three unstretched springs of spring constant k, 2k, 3k are attached to cylinder as shown in the figure. Find the period of small oscillations. Given that surface is rough enough to prevent slipping of cylinder. Attached the picture in the replies. Also, the answer is supposed to be 2*pi*sqrt(3m/23k)
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The period of small oscillations is T = 2π √3m / 23K
Explanation:
- Let there is sufficient friction to support pure rolling
- Let the cylinder rotates by an angle θ clockwise.
- Compression in the upper spring=(2R)×(θ)
- Compression in the middle spring = (R+R2)×(θ)=3Rθ2
- Compression in the lower spring = R×θ
Torque applied by the Upper spring in anticlockwise is = Force × lever arm=(k)×(2Rθ)×(2R) = 4KR^2θ
Torque applied by the middle spring in anticlockwise is = Force × lever arm=(2.k)×(3 /2 Rθ)×(3/2 R) = 9 / 2 KR^2θ
Torque applied by the lower spring in anticlockwise is =Force ×lever arm=(3k) × (Rθ) × (R)=3 K R^2θ
Net anticlockwise torque = 4KR^2θ + 9 / 2KR^2θ + 3KR^2θ = (7 + 92) =
23 / 2KR^2θ
Iα = 23/2KR^2θ
(mR^2 / 2 + mR^2) α = 23/2 KR^2θ
3/2 mR^2 × α = 23 /2KR^2 θ
3 m × α = 23Kθ
α = 23Kθ / 3 m
ω^2θ = 23Kθ / 3.m
ω^2 = 23 K/3.m
ω = √23K / 3.m
2π/T = √23 K / 3m
T = 2π √3m / 23K
Hence the period of small oscillations is T = 2π √3m / 23K
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