
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
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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, ${K}^{2}= \dfrac {2}{5} {R}^{2}$.
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