Question

A rigid body starts from rest and then rolls down a plane inclined at 11° to the horizontal. If the ratio of the radius of

gyration to the radius of the body is 0.6 and if g is 10 m/s². What is the linear acceleration of the body while rolling

down the plane?​

Answer 1

Given:

A rigid body starts from rest and then rolls down a plane inclined at 11° to the horizontal. If the ratio of the radius of gyration to the radius of the body is 0.6 and if g is 10 m/s².

To find:

Linear acceleration of the body?

Calculation:

For any body rolling down an inclined plane, thre are 3 equations:

• $mg\sin(\theta)-f=ma$
• fR = mk²$\alpha$
• a = $R\alpha$

When we solve these three equations, we get:

$\boxed{ \bf \: a = \dfrac{g \sin( \theta) }{ \bigg(1 + \dfrac{ {k}^{2} }{ {R}^{2} } \bigg)} }$

Now, put the values:

$\rm \implies \: a = \dfrac{g \sin( \theta) }{ \bigg \{1 + \bigg({\dfrac{k}{R} \bigg)}^{2} \bigg \}}$

$\rm \implies \: a = \dfrac{g \sin( {11}^{ \circ} ) }{ \bigg \{1 + \bigg({0.6\bigg)}^{2} \bigg \}}$

$\rm \implies \: a = \dfrac{10 \times \sin( {11}^{ \circ} ) }{ \bigg \{1 + \bigg({0.6\bigg)}^{2} \bigg \}}$

$\rm \implies \: a = \dfrac{10 \times \sin( {11}^{ \circ} ) }{ \bigg \{1 + 0.36 \bigg \}}$

$\rm \implies \: a = 7.34 \times \sin( {11}^{ \circ} )$

$\rm \implies \: a = 1.4 \: m/ {s}^{2}$

So, linear acceleration is 1.4 m/s².