A cylinder of length 1.5m and diameter 4 cm is fixed at one end .A tangential force of 4×10^5N is applied at the other end If the rigidity modulus of the cylinder is 6×10^10Nm-2 then calculate the twist produced in the cylinder ?

Question

Answer

Answer

Twist produced in a cylinder

Given data:

Length of the cylinder (L) = 1.5 m

Diameter of the cylinder (D) = 4 cm

Tangential force applied (F) = 4 × 10^5 N

Rigidity modulus of the cylinder (G) = 6 × 10^10 N/m^2

Calculation:

Step 1: Finding the radius of the cylinder

The diameter of the cylinder is given as 4 cm. To find the radius, we divide the diameter by 2.
Radius (r) = D/2 = 4 cm / 2 = 2 cm = 0.02 m

Step 2: Calculating the torque applied

The torque (τ) applied to the cylinder can be calculated using the formula:
τ = F × r
Substituting the given values:
τ = (4 × 10^5 N) × (0.02 m)
τ = 8 × 10^3 Nm

Step 3: Calculating the twist angle

The twist angle (θ) can be calculated using the formula:
θ = τ / (G × L)
Substituting the given values:
θ = (8 × 10^3 Nm) / (6 × 10^10 N/m^2 × 1.5 m)
θ = 8 × 10^3 Nm / 9 × 10^10 Nm
θ = 8 / 9 × 10^-7 rad

Result:

The twist produced in the cylinder is approximately 8 / 9 × 10^-7 rad.

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