
The wheel is making revolutions about its axis with uniform angular acceleration. Starting from rest, it reaches $100 rev/sec$ in $4 seconds$. Find the angular acceleration. Find the angle rotated during these four seconds.
Answer
574.2k+ views
Hint
We should know that the angular acceleration is the time rate of change of the angular velocity and is usually that is shown by alpha and is expressed in radians per second. Based on this concept we have to solve this question.
Complete step-by step answer
Let us consider the angular acceleration of the wheel to be $\alpha$.
Then initial angular velocity ($\omega _0$) = 0 and final angular velocity = $\omega$
$\omega = 100 \times 2\pi \;\dfrac{{rad}}{s} = 200\pi \;\dfrac{{rad}}{s}(As\;\omega = 100\;\dfrac{{rev}}{s})$
Now applying equations, we get:
$\omega = {\omega _0} + \alpha t$
$\Rightarrow 200\pi \;\dfrac{{rad}}{s} = 0 + \alpha (4)(t = 4s)$
$\Rightarrow \alpha = \dfrac{{200\pi }}{4} = \dfrac{{50\pi rad}}{{{s^2}}}$
The angle rotated in this time t is given by:
$\theta = {\omega _0}t + \dfrac{1}{2}a{t^2}$
$\theta = 0 + \dfrac{1}{2} \times 50 \times {4^2} = 25 \times 16 = 400\pi \;radians$
Hence, the answer is $400\pi \;radians$.
Note
The angular acceleration is also known as rotational acceleration and the expression that is formed is considered to be as a vector quantity. It is considered a vector quantity because it consists of a magnitude component and either of two defined directions or we can sense. At every point of a rigid body we find that there is the presence of rotational velocity and acceleration.
We should know that the angular acceleration is the time rate of change of the angular velocity and is usually that is shown by alpha and is expressed in radians per second. Based on this concept we have to solve this question.
Complete step-by step answer
Let us consider the angular acceleration of the wheel to be $\alpha$.
Then initial angular velocity ($\omega _0$) = 0 and final angular velocity = $\omega$
$\omega = 100 \times 2\pi \;\dfrac{{rad}}{s} = 200\pi \;\dfrac{{rad}}{s}(As\;\omega = 100\;\dfrac{{rev}}{s})$
Now applying equations, we get:
$\omega = {\omega _0} + \alpha t$
$\Rightarrow 200\pi \;\dfrac{{rad}}{s} = 0 + \alpha (4)(t = 4s)$
$\Rightarrow \alpha = \dfrac{{200\pi }}{4} = \dfrac{{50\pi rad}}{{{s^2}}}$
The angle rotated in this time t is given by:
$\theta = {\omega _0}t + \dfrac{1}{2}a{t^2}$
$\theta = 0 + \dfrac{1}{2} \times 50 \times {4^2} = 25 \times 16 = 400\pi \;radians$
Hence, the answer is $400\pi \;radians$.
Note
The angular acceleration is also known as rotational acceleration and the expression that is formed is considered to be as a vector quantity. It is considered a vector quantity because it consists of a magnitude component and either of two defined directions or we can sense. At every point of a rigid body we find that there is the presence of rotational velocity and acceleration.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Which cell organelles are present in white blood C class 11 biology CBSE

What is the molecular geometry of BrF4 A square planar class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

State the laws of reflection of light

Difference Between Prokaryotic Cells and Eukaryotic Cells

Show that total energy of a freely falling body remains class 11 physics CBSE

