Question

A stationary wheel starts rotating about its own axis at uniform angular acceleration 8rad/sec2 .The time taken by t to comlete 77 rotations is

Answer 1

Θ = (ωi × t) + 0.5αt²
It was initially at rest. Hence ωi = 0

θ = 0.5αt²

t = √(2θ / α)
= √(2 × (77 × 2 π rad) / 8 rad/s²)
= 11 seconds ………[Here I took π = 22/7]

Time taken to complete 77 revolutions is 11 seconds

[Note: We should multiply revolutions by 2 π to

Answer 2

Question:

• A stationary wheel starts rotating about its own axis at uniform angular acceleration 8rad/sec2 .The time taken by t to complete 77 rotations is

Answer:

• The time taken to complete 77 rotations is 11 seconds

Explanation:

Given that:

• Angular displacement is 2π × 77 rad
• initial angular velocity is 0
• Angular acceleration is 8 rad / sec²

To Find:

• The time taken by the wheel to complete 77 rotations

Required Solution:

• Using formula to find the angular displacement

$\leadsto {\pink{\boxed{\bf{ \emptyset = \omega_0 t + \frac{1}{2} \alpha t^2 }}}}$

Where ,

• ∅ denotes angular displacement
• ω₀ denotes initial angular velocity
• α denotes angular acceleration
• t denotes the time period

Calculations :

$\longrightarrow \rm \emptyset = \omega_0 t + \dfrac{1}{2} \alpha t^2$

$\longrightarrow \rm 2\pi * 77 = 0 * t + \dfrac{1}{2} * 8 * t^2$

$\longrightarrow \rm 2\pi * 77 = \dfrac{1}{2} * 8 * t^2$

$\longrightarrow \rm 2\pi * 77 = 4 * t^2$

$\longrightarrow \rm 2 * 22/7 * 77 = 4 * t^2$

$\longrightarrow \rm 44 * 11 = 4 * t^2$

$\longrightarrow \rm 11 * 11 = t^2$

$\longrightarrow \rm t = \sqrt{11 * 11}$

$\longrightarrow {\red{\underline{\underline{\rm{ Time _{(to \; complete \; 77 \; rotations)} = 11 \; seconds }}}}}$

$\qquad {\underline{\pmb{\sf{Henceforth , the \; time \; taken \; by \; the \; wheel \; is \; 11 \; seconds }}}}$

Therefore:

• The time taken to complete 77 rotations is 11 seconds

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