
Two identical uniform discs of mass m and radius r are arranged as shown in the figure. If \[\alpha \] is the angular acceleration of the lower disc and is the acceleration of the center of mass of the lower disc, then find the relation among \[{a_{cm}}\], \[\alpha \] and r.

A. \[{a_{cm}} = \dfrac{\alpha }{r}\]
B. \[{a_{cm}} = 2\alpha r\]
C. \[{a_{cm}} = \alpha r\]
D. None of these
Answer
163.8k+ views
Hint:Before we start addressing the problem, we need to know about what data has been provided and what we need to solve. Here they have given the mass of two identical discs as m and r , the angular acceleration of the lower disc is \[\alpha \] and the acceleration of the center of mass of the lower disc is \[{a_{cm}}\], then we need to find the relation among \[{a_{cm}}\], \[\alpha \] and r.
Complete step by step solution:

Image: Two identical uniform discs.
As shown in figure the tension T is acting on the discs having radius r
If we apply torque for the above disc at the centre then,
\[\tau = I \times \alpha \]
\[T \times r = \dfrac{{m{r^2}}}{2} \times {\alpha _1}\]
\[{\alpha _1} = \dfrac{{2T}}{{mr}}\]…… (1)
Now, if we apply the torque to the lower disc we get,
\[T \times r = \dfrac{{m{r^2}}}{2} \times \alpha \]
\[\alpha = \dfrac{{2T}}{{mr}}\]……….. (2)
From equation (1) and (2) we will have,
\[{\alpha _1} = \alpha \]….. (1)
Since the centre of mass of the lower disc passes through its centre then the centre of mass will become zero. Then, the acceleration of the above disc = Acceleration of the below disc
\[r{\alpha _1} = {a_{cm}} - r\alpha \]
Substitute the value \[{\alpha _1} = \alpha \] in above equation we get,
\[{a_{cm}} = 2r\alpha \]
Therefore, then the relation among \[{a_{cm}}\], \[\alpha \] and r is \[{a_{cm}} = 2r\alpha \].
Hence, option B is the correct answer.
Note: Here, in this problem it is important to remember that, the definition of angular acceleration, and also when the two identical uniform discs of mass m and radius r are connected through a string as shown we need to identify in which direction the torque will be acting.
Complete step by step solution:

Image: Two identical uniform discs.
As shown in figure the tension T is acting on the discs having radius r
If we apply torque for the above disc at the centre then,
\[\tau = I \times \alpha \]
\[T \times r = \dfrac{{m{r^2}}}{2} \times {\alpha _1}\]
\[{\alpha _1} = \dfrac{{2T}}{{mr}}\]…… (1)
Now, if we apply the torque to the lower disc we get,
\[T \times r = \dfrac{{m{r^2}}}{2} \times \alpha \]
\[\alpha = \dfrac{{2T}}{{mr}}\]……….. (2)
From equation (1) and (2) we will have,
\[{\alpha _1} = \alpha \]….. (1)
Since the centre of mass of the lower disc passes through its centre then the centre of mass will become zero. Then, the acceleration of the above disc = Acceleration of the below disc
\[r{\alpha _1} = {a_{cm}} - r\alpha \]
Substitute the value \[{\alpha _1} = \alpha \] in above equation we get,
\[{a_{cm}} = 2r\alpha \]
Therefore, then the relation among \[{a_{cm}}\], \[\alpha \] and r is \[{a_{cm}} = 2r\alpha \].
Hence, option B is the correct answer.
Note: Here, in this problem it is important to remember that, the definition of angular acceleration, and also when the two identical uniform discs of mass m and radius r are connected through a string as shown we need to identify in which direction the torque will be acting.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
