What is the moment of inertia for a solid sphere w.r.t. a tangent touching to its surface?

Question

What is the moment of inertia for a solid sphere w.r.t. a tangent touching to its surface? (A) (2/5)MR2 (B) (7/5)MR2 (C) (2/3)MR2 (D) (5/3)MR2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c70e36b6e042bf0f52149fec2c7db60f-.png" /> The moment of inertia for a solid sphere along its diameter is I text diameter =(2/5) M R2 Moment of inertia about a tangent touching to its surface, I text tangent =I text diameter +M R2 (using theorem of parallel axes) =(2/5) M R2+M R2 =(7/5) M R2

Q. What is the moment of inertia for a solid sphere w.r.t. a tangent touching to its surface?

$\frac{2}{5} M R^{2}$

$\frac{7}{5} M R^{2}$

$\frac{2}{3} M R^{2}$

$\frac{5}{3} M R^{2}$

The moment of inertia for a solid sphere along its diameter is
$I_{diameter} = \frac{2}{5} M R^{2}$
Moment of inertia about a tangent touching to its surface,
$I_{tangent} = I_{diameter} + M R^{2}$
(using theorem of parallel axes)
$= \frac{2}{5} M R^{2} + M R^{2}$
$= \frac{7}{5} M R^{2}$

Q. What is the moment of inertia for a solid sphere w.r.t. a tangent touching to its surface?

52​MR2

57​MR2

32​MR2

35​MR2

The moment of inertia for a solid sphere along its diameter is Idiameter ​=52​MR2 Moment of inertia about a tangent touching to its surface, Itangent ​=Idiameter ​+MR2 (using theorem of parallel axes) =52​MR2+MR2 =57​MR2

$\frac{2}{5} M R^{2}$

$\frac{7}{5} M R^{2}$

$\frac{2}{3} M R^{2}$

$\frac{5}{3} M R^{2}$

The moment of inertia for a solid sphere along its diameter is
$I_{diameter} = \frac{2}{5} M R^{2}$
Moment of inertia about a tangent touching to its surface,
$I_{tangent} = I_{diameter} + M R^{2}$
(using theorem of parallel axes)
$= \frac{2}{5} M R^{2} + M R^{2}$
$= \frac{7}{5} M R^{2}$