Question 505: Centre of Mass and Rotational Motion

Question

An inclined plane makes an angle $30^{\circ}$ with the horizontal. A solid sphere rolling down an inclined plane from rest without slipping has linear acceleration ( $\mathrm{g}=$ acceleration due gravity) ( $\sin 30^{\circ}=0.5$ )

Accepted Answer

For a solid sphere rolling down an inclined plane
without slipping
:
Let the
angle of inclination
be $	heta = 30^\circ$
Let the
linear acceleration
be $a$.
Let the
radius
of the sphere be $R$ and its
mass
be $m$.
Step 1: Forces Acting on Sphere
Down the incline: $mg\sin	heta$ (component of weight)
Up the incline: Friction
Step 2: Equations of Motion
Translational motion (along incline):
$$ mg\sin	heta - f = ma $$
Where $f$ is the frictional force.
Rotational motion (about the centre of mass):
$$ fR = I\alpha $$
Where $I$ is the moment of inertia about its centre, and $\alpha$ is the angular acceleration.
For a
solid sphere
,
$$ I = \frac{2}{5}mR^2 $$
Step 3: Rolling Condition
Since the sphere rolls
without slipping
, velocity relation:
$$ a = R\alpha $$
So,
$$ fR = I\alpha = I\left(\frac{a}{R}ight) = \frac{2}{5}mR^2 \cdot \frac{a}{R} = \frac{2}{5}mRa $$
Or,
$$ f = \frac{2}{5}ma $$
Step 4: Substitute $f$ in the Translational Equation
Recall:
$$ mg\sin	heta - f = ma $$
Plug in $f$:
$$ mg\sin	heta - \frac{2}{5}ma = ma $$
So,
$$ mg\sin	heta = ma + \frac{2}{5}ma = ma \left(1 + \frac{2}{5}ight) = ma \cdot \frac{7}{5} $$
Step 5: Solve for $a$
$$ mg\sin	heta = \frac{7}{5}ma $$
$$ a = \frac{5}{7}g\sin	heta $$
Step 6: Plug in $\sin 30^\circ = 0.5$
$$ a = \frac{5}{7}g 	imes 0.5 = \frac{5g}{14} $$
Final Answer
Option B
is correct.
$$ \boxed{\frac{5g}{14}} $$

Answer

$\frac{5 \mathrm{~g}}{7}$

Answer

$\frac{5 \mathrm{~g}}{14}$

Answer

$\frac{2 \mathrm{~g}}{3}$

Answer

$\frac{\mathrm{g}}{3}$

Centre of Mass and Rotational Motion

An inclined plane makes an angle 30^{\circ} with the horizontal. A solid sphere rolling down an inclined plane from rest without slipping has linear acceleration ( \mathrm{g}= acceleration due gravity) ( \sin 30^{\circ}=0.5 )

Choose the correct answer: