Rotational Motion

For equilibrium, the sum of vertical and horizontal forces must be zero, and the sum of torques about any point must also be zero.$

%5Cbegin%7Baligned%7D%20%26%5Ctext%20%7B%20From%20the%20diagram%2C%20we%20can%20see%20that%3A%20%7D%5C%5C%20%26N_2%2Bf_1%3Dm%20g%20%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2CN_1%3Df_2%3D%5Cmu_2%20N_2%5Cend%7Baligned%7D

%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%20%5Cbegin%7Baligned%7D%20%26%20f_1%3D%5Cmu_1%20N_1%3D%5Cmu_1%20%5Cmu_2%20N_2%20%5C%5C%20%26%20N_2%5Cleft(1%2B%5Cmu_1%20%5Cmu_2%5Cright)%3Dm%20g%20%5C%5C%20%26%20%5CRightarrow%20N_2%3D%5Cfrac%7Bm%20g%7D%7B%5Cleft(1%2B%5Cmu_1%20%5Cmu_2%5Cright)%7D%20%5Cend%7Baligned%7D

Torque%20Balance%3ABalancing%20the%20torque%2C%20we%20get%2C

f_1%20l%20%5Ccos%20(%5Calpha)%2BN_1%20l%20%5Csin%20(%5Calpha)%3Dm%20g%20%5Cfrac%7Bl%7D%7B2%7D%20%5Ccos%20(%5Calpha)

where

is%20the%20length%20of%20the%20rod.

%5Cbegin%7Bgathered%7D%20f_1%2BN_1%20%5Ctan%20(%5Calpha)%3D%5Cfrac%7Bm%20g%7D%7B2%7D%20%5C%5C%20N_1%20%5Ctan%20(%5Calpha)%3D%5Cfrac%7Bm%20g%7D%7B2%7D-f_1%3D%5Cfrac%7Bm%20g%7D%7B2%7D-%5Cmu_1%20N_1%20%5Cend%7Bgathered%7D

%5Cbegin%7Baligned%7D%20%26%5CRightarrow%20%5Ctan%20(%5Calpha)%3D%5Cleft(%5Cfrac%7Bm%20g%7D%7B2%20N_1%7D-%5Cmu_1%5Cright)%5C%5C%20%26%5Ctext%20%7B%20So%2C%20the%20angle%20of%20inclination%20%7D%20%5Calpha%20%5Ctext%20%7B%2C%20such%20that%20the%20rod%20stays%20in%20equilibrium%20is%20%7D%5C%5C%20%26%5Cbegin%7Bgathered%7D%20%5Calpha%3D%5Ctan%20%5E%7B-1%7D%5Cleft(%5Cfrac%7B1%2B%5Cmu_1%20%5Cmu_2%7D%7B2%20%5Cmu_2%7D-%5Cmu_1%5Cright)%20%5C%5C%20%5Calpha%3D%5Ctan%20%5E%7B-1%7D%5Cleft(%5Cfrac%7B1-%5Cmu_1%20%5Cmu_2%7D%7B2%20%5Cmu_2%7D%5Cright)%20%5Cend%7Bgathered%7D%20%5Cend%7Baligned%7D

For the system described, two blocks of masses

M

and

M%2F2

are connected by an inextensible cord over a solid disc, and we're tasked with finding the angular acceleration of the disc. Let's go through the process:Tensions in the cord:For mass

M

M%20g%20-%20T_1%20%3D%20M%20a%20%5Cquad%20%5CRightarrow%20%5Cquad%20T_1%20%3D%20M(g%20-%20a)%20%5Cquad%20%5Ctext%7B(Equation%201)%7D

For mass

M%2F2

T_2%20-%20%5Cfrac%7BM%7D%7B2%7D%20g%20%3D%20%5Cfrac%7BM%7D%7B2%7D%20a%20%5Cquad%20%5CRightarrow%20%5Cquad%20T_2%20%3D%20%5Cfrac%7BM%7D%7B2%7D(g%20-%20a)%20%5Cquad%20%5Ctext%7B(Equation%202)%7D

Relation between linear and angular accelerations:Since the cord is inextensible and there's no slipping, the linear acceleration

a

of the blocks relates to the angular acceleration

%5Calpha

of the disc by:

a%20%3D%20R%20%5Calpha

Torque on the disc:The torque due to the tensions on the disc is given by:

%5Ctext%7BTorque%7D%20%3D%20T_1%20R%20-%20T_2%20R%20%3D%20I%20%5Calpha

The moment of inertia

I

for the disc rotating about its center is:

I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20M%20R%5E2

Thus,:

(T_1%20-%20T_2)%20R%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20M%20R%5E2%20%5Calpha%20%5Cquad%20%5Ctext%7B(Equation%203)%7D

Substitute tensions into the torque equation:Substituting Equations 1 and 2 into Equation 3, we have:

M(g%20-%20R%20%5Calpha)%20-%20%5Cfrac%7BM%7D%7B2%7D(R%20%5Calpha%20%2B%20g)%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20M%20R%20%5Calpha

Simplify and solve for

%5Calpha

:Simplify the equation:

M%20g%20-%20M%20R%20%5Calpha%20-%20%5Cfrac%7BM%7D%7B2%7D%20R%20%5Calpha%20-%20%5Cfrac%7BM%7D%7B2%7D%20g%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20M%20R%20%5Calpha

Arrange and combine terms:

2g%20-%202R%20%5Calpha%20-%20(R%20%5Calpha%20%2B%20g)%20%3D%20R%20%5Calpha

Simplify further:

2g%20-%202R%20%5Calpha%20-%20R%20%5Calpha%20-%20g%20%3D%20R%20%5Calpha

g%20%3D%204R%20%5Calpha

Thus, the angular acceleration

%5Calpha

is:

%5Calpha%20%3D%20%5Cfrac%7Bg%7D%7B4R%7D

About Rotational Motion - IAT-IISER

Rotational Motion is a vital chapter for IAT-IISER 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 Rotational 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 Rotational Motion carry the most weight. Then, tackle the questions iteratively to solidify your understanding.

High-Yield Trend

Chapter Questions
3 MCQs

Official Solution

Official Solution

Official Solution

About Rotational Motion - IAT-IISER

Frequently Asked Questions

Why focus on Rotational Motion PYQs?

How to best use this analysis?

Rotational Motion

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

About Rotational Motion - IAT-IISER

Frequently Asked Questions

Why focus on Rotational Motion PYQs?

How to best use this analysis?

Chapter Questions
3 MCQs