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IV. Exercises · Q2

Q.A cylinder of length 1.5 m and diameter 4 cm is fixed at one end. A tangential force of 4×1054\times10^{5} N is applied at the other end. If the rigidity modulus of the cylinder is 6×10106\times10^{10} N m−2^{-2} then, calculate the twist produced in the cylinder.

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✓ Free question

Step 1. The cylinder has length L=1.5L=1.5 m and diameter 4 cm, so radius r=0.02r=0.02 m; a tangential force F=4×105F=4\times10^5 N is applied at the rim of the free end, and the rigidity modulus is ηR=6×1010\eta_R=6\times10^{10} N m−2^{-2}.

Step 2. The torque produced by the tangential force about the cylinder's axis is τ=Fr=(4×105)(0.02)=8000\tau=Fr=(4\times10^5)(0.02)=8000 N m.

Step 3. The angle of twist of the free end of a solid cylindrical rod, rigidly fixed at the other end, is θ=2τLπr4ηR\theta=\dfrac{2\tau L}{\pi r^4\eta_R}.

Step 4. Compute r4=(0.02)4=1.6×10−7r^4=(0.02)^4=1.6\times10^{-7} m4^4; then πr4ηR=π×1.6×10−7×6×1010=π×9600≈30159\pi r^4\eta_R=\pi\times1.6\times10^{-7}\times6\times10^{10}=\pi\times9600\approx30159.

Step 5. θ=2×8000×1.530159=2400030159≈0.7958\theta=\dfrac{2\times8000\times1.5}{30159}=\dfrac{24000}{30159}\approx0.7958 rad.

Step 6. Converting to degrees: θ≈0.7958×180π≈45.6∘\theta\approx0.7958\times\dfrac{180}{\pi}\approx45.6^{\circ}.

✓Final answer

The twist produced in the cylinder is about 45.6∘45.6^{\circ}.

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