Uniform Circular Motion

The net acceleration $a_%7B%5Ctext%7Bnet%7D%7D$ can be expressed as the vector sum of the tangential acceleration $a_t$ and the centripetal acceleration $a_c$ . The centripetal acceleration is given by: $a_c%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D$ Where:
- $v$ is the instantaneous speed of the ball,
- $r%20%3D%205%20%5C%2C%20%5Ctext%7Bm%7D$ is the radius. Given:
- The tangential acceleration $a_t%20%3D%2010%20%5C%2C%20%5Ctext%7Bm%2Fs%7D%5E2$ ,
- The angle between the net acceleration and the centripetal acceleration is $30%5E%5Ccirc$ . Using the relation for the net acceleration: $a_%7B%5Ctext%7Bnet%7D%7D%5E2%20%3D%20a_c%5E2%20%2B%20a_t%5E2$ Also, the net acceleration is related to the angle $%5Ctheta$ between the tangential and centripetal accelerations: $a_%7B%5Ctext%7Bnet%7D%7D%20%3D%20%5Csqrt%7Ba_t%5E2%20%2B%20a_c%5E2%7D$ Since $%5Ctheta%20%3D%2030%5E%5Ccirc$ , the tangent of the angle gives: $%5Ctan(30%5E%5Ccirc)%20%3D%20%5Cfrac%7Ba_t%7D%7Ba_c%7D$ This gives: $%5Cfrac%7B10%7D%7Ba_c%7D%20%3D%20%5Ctan(30%5E%5Ccirc)%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B3%7D%7D$ Solving for $a_c$ : $a_c%20%3D%2010%5Csqrt%7B3%7D%20%5C%2C%20%5Ctext%7Bm%2Fs%7D%5E2$ Now substitute the value of $a_c$ into the centripetal acceleration formula: $a_c%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20%5CRightarrow%2010%5Csqrt%7B3%7D%20%3D%20%5Cfrac%7Bv%5E2%7D%7B5%7D$ Solving for $v$ : $v%5E2%20%3D%2050%5Csqrt%7B3%7D$ $v%20%3D%20%5Csqrt%7B50%5Csqrt%7B3%7D%7D%20%5Capprox%209.3%20%5C%2C%20%5Ctext%7Bm%2Fs%7D$
Thus, the instantaneous speed is $9.3%20%5C%2C%20%5Ctext%7Bm%2Fs%7D$ .

The centripetal acceleration

a

for a body moving in a circular path is given by the formula:

a%20%3D%20%5Comega%5E2%20r

Where: -

%5Comega

is the angular velocity, and -

r

is the radius of the circular path. The angular velocity

%5Comega

is related to the frequency of revolution

f

by the relation:

%5Comega%20%3D%202%5Cpi%20f

Given that the frequency of revolution

f

is numerically equal to the radius of the circular path

r

, we can write:

f%20%3D%20r

So, the angular velocity becomes:

%5Comega%20%3D%202%5Cpi%20r

Substituting this into the formula for acceleration:

a%20%3D%20(2%5Cpi%20r)%5E2%20r%20%3D%204%5Cpi%5E2%20r%5E3

Now, substitute the given radius

r%20%3D%20%5Cfrac%7B5%7D%7B%5Cpi%7D%20%5C%2C%20%5Ctext%7Bm%7D

a%20%3D%204%5Cpi%5E2%20%5Cleft(%20%5Cfrac%7B5%7D%7B%5Cpi%7D%20%5Cright)%5E3%20%3D%204%5Cpi%5E2%20%5Ctimes%20%5Cfrac%7B125%7D%7B%5Cpi%5E3%7D%20%3D%20%5Cfrac%7B500%7D%7B%5Cpi%7D%20%5C%2C%20%5Ctext%7Bms%7D%5E%7B-2%7D

Thus, the acceleration of the body is

%5Cleft(%20%5Cfrac%7B500%7D%7B%5Cpi%7D%20%5Cright)%20%5C%2C%20%5Ctext%7Bms%7D%5E%7B-2%7D

About Uniform Circular Motion - COMEDK-UGET

Uniform Circular Motion is a vital chapter for COMEDK-UGET aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.

By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.

Frequently Asked Questions

Why focus on Uniform Circular Motion PYQs?

Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.

How to best use this analysis?

Review the topic breakdown to see which sub-topics within Uniform Circular Motion carry the most weight. Then, tackle the questions iteratively to solidify your understanding.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

About Uniform Circular Motion - COMEDK-UGET

Frequently Asked Questions

Why focus on Uniform Circular Motion PYQs?

How to best use this analysis?

Uniform Circular Motion

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

About Uniform Circular Motion - COMEDK-UGET

Frequently Asked Questions

Why focus on Uniform Circular Motion PYQs?

How to best use this analysis?

Chapter Questions
2 MCQs