KCET · Physics · Rotational Motion
A wheel starting from rest gains an angular velocity of \(10 \mathrm{rad} / \mathrm{s}\) after uniformly accelerated for \(5 \mathrm{~s}\). The total angle through which it has turned is
- A \(25 \mathrm{rad}\)
- B \(100 \mathrm{rad}\)
- C \(25 \pi \mathrm{rad}\)
- D \(50 \pi \mathrm{rad}\) and a vertical axis
Answer & Solution
Correct Answer
(A) \(25 \mathrm{rad}\)
Step-by-step Solution
Detailed explanation
Initial angular velocity of wheel, \(\omega_{0}=0\)
Final angular velocity, \(\omega=10 \mathrm{rad} / \mathrm{s}\)
\(t=5 \mathrm{~s}\)
By first equation of rotational motion,
\(\omega=\omega_{0}+\alpha t \)
\( \Rightarrow 10=0+\alpha \times 5 \)
\( \Rightarrow \alpha=\frac{10}{5}=2 \mathrm{rad} / \mathrm{s}^{2}\)
If \(\theta\) be the total angle through which wheel has turned, then from,
\(\theta =\omega_{0} t+\frac{1}{2} \alpha t^{2} \)
\( =0 \times 5+\frac{1}{2} \times 2 \times 5^{2}=0+25 \)
\( \Rightarrow \theta =25 \mathrm{rad}\)
Final angular velocity, \(\omega=10 \mathrm{rad} / \mathrm{s}\)
\(t=5 \mathrm{~s}\)
By first equation of rotational motion,
\(\omega=\omega_{0}+\alpha t \)
\( \Rightarrow 10=0+\alpha \times 5 \)
\( \Rightarrow \alpha=\frac{10}{5}=2 \mathrm{rad} / \mathrm{s}^{2}\)
If \(\theta\) be the total angle through which wheel has turned, then from,
\(\theta =\omega_{0} t+\frac{1}{2} \alpha t^{2} \)
\( =0 \times 5+\frac{1}{2} \times 2 \times 5^{2}=0+25 \)
\( \Rightarrow \theta =25 \mathrm{rad}\)
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