
The ratio of the acceleration for a solid sphere (mass ‘m’ and radius ‘R’) rolling down an incline of angle $’\theta’$ without slipping and slipping down the incline without rolling is.
A. 5:7
B. 2:3
C. 2:5
D. 7:5
Answer
564.9k+ views
Hint: To solve this problem, find the acceleration of the solid sphere rolling down an incline without slipping. Substitute the value of ratio of radius of gyration to radius of the sphere. Solve it and find the acceleration of the sphere. Then, find the acceleration of the solid sphere slipping down an incline without rolling. Divide both the equations of acceleration to get the ratio of their accelerations.
Formula used: ${K}^{2}=\dfrac {2}{5} {R}^{2}$
Complete step by step answer:
Acceleration of the solid sphere rolling down an incline without slipping is given by,
${a}_{roll}= \dfrac{g \sin {\theta}}{1+ \dfrac {{K}^{2}}{{r}^{2}}}$ …(1)
Where, K is the radius of gyration
r is the radius of the solid sphere
For a solid sphere,
${K}^{2}=\dfrac {2}{5} {R}^{2}$
$\Rightarrow \dfrac {{K}^{2}}{{r}^{2}}= \dfrac {2}{5}$
Substituting this value in the equation. (1) we get,
${a}_{roll}= \dfrac {g \sin {\theta}}{1+ \dfrac {2}{5}}$
$\Rightarrow {a}_{roll}= \dfrac {g\ sin {\theta}}{\dfrac {7}{5}}$
$\Rightarrow {a}_{roll}= \dfrac {5}{7}g \sin {\theta}$ …(2)
Acceleration of the solid sphere slipping down an incline without rolling is given by,
${a}_{slip}= g \sin {\theta}$ …(3)
Dividing equation. (2) by equation. (3) we get,
$\dfrac {{a}_{roll}}{{a}_{slip}}= \dfrac {5}{7}$
$\Rightarrow {a}_{roll}:{a}_{slip}= 5:7$
Hence, the ratio of acceleration for a solid sphere rolling down an incline without slipping and slipping down the incline without rolling is 5:7.
So, the correct answer is “Option A”.
Note: To solve these types of questions, students must know the relationship between radius of gyration (K) and radius (R) of a cylinder, hollow cylinder, sphere and hollow sphere. For a cylinder, the relationship between K and R is given by, ${K}^{2}= \dfrac {1}{2} {R}^{2}$ while for a hollow cylinder it is given by, ${K}^{2}= {R}^{2}$. For a hollow sphere it is given by, ${K}^{2}= \dfrac {2}{3} {R}^{2}$ and for a sphere it is given by, ${K}^{2}= \dfrac {2}{5} {R}^{2}$.
Formula used: ${K}^{2}=\dfrac {2}{5} {R}^{2}$
Complete step by step answer:
Acceleration of the solid sphere rolling down an incline without slipping is given by,
${a}_{roll}= \dfrac{g \sin {\theta}}{1+ \dfrac {{K}^{2}}{{r}^{2}}}$ …(1)
Where, K is the radius of gyration
r is the radius of the solid sphere
For a solid sphere,
${K}^{2}=\dfrac {2}{5} {R}^{2}$
$\Rightarrow \dfrac {{K}^{2}}{{r}^{2}}= \dfrac {2}{5}$
Substituting this value in the equation. (1) we get,
${a}_{roll}= \dfrac {g \sin {\theta}}{1+ \dfrac {2}{5}}$
$\Rightarrow {a}_{roll}= \dfrac {g\ sin {\theta}}{\dfrac {7}{5}}$
$\Rightarrow {a}_{roll}= \dfrac {5}{7}g \sin {\theta}$ …(2)
Acceleration of the solid sphere slipping down an incline without rolling is given by,
${a}_{slip}= g \sin {\theta}$ …(3)
Dividing equation. (2) by equation. (3) we get,
$\dfrac {{a}_{roll}}{{a}_{slip}}= \dfrac {5}{7}$
$\Rightarrow {a}_{roll}:{a}_{slip}= 5:7$
Hence, the ratio of acceleration for a solid sphere rolling down an incline without slipping and slipping down the incline without rolling is 5:7.
So, the correct answer is “Option A”.
Note: To solve these types of questions, students must know the relationship between radius of gyration (K) and radius (R) of a cylinder, hollow cylinder, sphere and hollow sphere. For a cylinder, the relationship between K and R is given by, ${K}^{2}= \dfrac {1}{2} {R}^{2}$ while for a hollow cylinder it is given by, ${K}^{2}= {R}^{2}$. For a hollow sphere it is given by, ${K}^{2}= \dfrac {2}{3} {R}^{2}$ and for a sphere it is given by, ${K}^{2}= \dfrac {2}{5} {R}^{2}$.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

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

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

