7.18 A solid sphere rolls down two different inclined planes of the same heights but different angles of inclination. (a) Will it reach the bottom with the same speed in each case? (b) Will it take longer to roll down one plane than the other? (c) If so, which one and why?

(a) Let the mass of the sphere = m

Height of the plane = h

Velocity of the sphere at the bottom of the plane = v

At the top of the plane, the total energy of the sphere = potential energy = mgh

At the bottom of the plane, the sphere has both translational and rotational energies.

Hence, total energy = (1/2)mv² + (1/2)I $\frac{-MR}{8}$

Using the law of conservation of energy, we can write: (1/2)mv²+ (1/2)I $\frac{4}{3M}$ = mgh …(1)

For a solid sphere, the moment of inertia, I = (2/5)mr²

The equation (1) becomes (1/2)mv² + (1/2)( (2/5)mr² $-\frac{R}{6}$ = mgh

(1/2) v² + (1/5)r² ${\omega }^2$ = gh

From the relation v = ${\omega }^2$ , we get

(1/2) v²+ (1/5) v²= gh

v = ${\omega }^2$

Since v depen

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(a) Let the mass of the sphere = m

Height of the plane = h

Velocity of the sphere at the bottom of the plane = v

At the top of the plane, the total energy of the sphere = potential energy = mgh

At the bottom of the plane, the sphere has both translational and rotational energies.

Hence, total energy = (1/2)mv² + (1/2)I $\frac{-MR}{8}$

Using the law of conservation of energy, we can write: (1/2)mv²+ (1/2)I $\frac{4}{3M}$ = mgh …(1)

For a solid sphere, the moment of inertia, I = (2/5)mr²

The equation (1) becomes (1/2)mv² + (1/2)( (2/5)mr² $-\frac{R}{6}$ = mgh

(1/2) v² + (1/5)r² ${\omega }^2$ = gh

From the relation v = ${\omega }^2$ , we get

(1/2) v²+ (1/5) v²= gh

v = ${\omega }^2$

Since v depends only on g (constant) and h (height = constant), the velocity at the bottom remains same from whichever plane the sphere rolled.

 

(b) Yes

 

(c)

Consider two inclined planes with inclinations ${\omega }^2$ and $r\omega$ related as $\surd \left(\frac{10}{7}\right)gh$ > ${\theta }_1$ .

The acceleration produced in the sphere where it rolls down the plane inclined at ${\theta }_2$ is given as g sin ${\theta }_2$ . Similarly for the plane with inclination ${\theta }_1$ is given as g sin ${\theta }_1$

Let R1 and R2 be the normal reaction to the sphere on these two planes

Since ${\theta }_1$ > ${\theta }_2$ , ${\theta }_2$ > sin ${\theta }_2$ ….(i)

${\theta }_1$ > ${sin\theta }_2$ …….(ii)

From the equation v = u +at where u = 0 and v = constant, we get t ${\theta }_1$ (1/a)

For inclination $a_2$ ,t₁ $a_1$ (1/a₁) and t₂$\alpha$ (1/a₂)

Therefore t₂ < t₁

So the sphere will take a longer time to reach plane with smaller inclination.

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<div><div><p><strong>(a)</strong> Let the mass of the sphere = m</p></div></div><p>Height of the plane = h</p><p>Velocity of the sphere at the bottom of the plane = v</p><p>At the top of the plane, the total energy of the sphere = potential energy = mgh</p><p>At the bottom of the plane, the sphere has both translational and rotational energies.</p><p>Hence, total energy = (1/2)mv² + (1/2)I <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mo>-</mo><mi>M</mi><mi>R</mi></mrow></mrow><mrow><mrow><mn>8</mn></mrow></mrow></mfrac></math></span></p><p>Using the law of conservation of energy, we can write: (1/2)mv²+ (1/2)I <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>4</mn></mrow></mrow><mrow><mrow><mn>3</mn><mi>M</mi></mrow></mrow></mfrac></math></span>&nbsp;= mgh …(1)</p><p>For a solid sphere, the moment of inertia, I = (2/5)mr²</p><p>The equation (1) becomes (1/2)mv² + (1/2)( (2/5)mr² <span title="Click to copy mathml"><math><mo>-</mo><mfrac><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>6</mn></mrow></mrow></mfrac></math></span>&nbsp;= mgh</p><p>(1/2) v² + (1/5)r² <span title="Click to copy mathml"><math><msup><mrow><mrow><mi>ω</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span>&nbsp;= gh</p><p>From the relation v =&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>ω</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span>&nbsp;, we get</p><p>(1/2) v²+ (1/5) v²= gh</p><p>v =&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>ω</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span></p><p>Since v depends only on g (constant) and h (height = constant), the velocity at the bottom remains same from whichever plane the sphere rolled.</p><p>&nbsp;</p><p><strong>(b)</strong> Yes</p><p>&nbsp;</p><p><strong>(c)</strong></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1729071853php4fCtf9_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1729071853php4fCtf9.jpeg" alt="" width="323" height="168"></picture></div></div><p>Consider two inclined planes with inclinations&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>ω</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span>&nbsp;and&nbsp;<span title="Click to copy mathml"><math><mi>r</mi><mi>ω</mi></math></span>&nbsp;related as&nbsp;<span title="Click to copy mathml"><math><mo>√</mo><mfenced separators="|"><mrow><mrow><mfrac><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>7</mn></mrow></mrow></mfrac></mrow></mrow></mfenced><mi>g</mi><mi>h</mi></math></span>&nbsp;&gt;&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span>&nbsp;.</p><p>The acceleration produced in the sphere where it rolls down the plane inclined at&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;is given as g sin&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;. Similarly for the plane with inclination&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span>&nbsp;is given as g sin&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span></p><p>Let R1 and R2 be the normal reaction to the sphere on these two planes</p><p>Since&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span>&nbsp;&gt;&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;&gt; sin&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;….(i)</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span>&nbsp;&gt;&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>s</mi><mi>i</mi><mi>n</mi><mi></mi><mi>θ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;…….(ii)</p><p>From the equation v = u +at where u = 0 and v = constant, we get t&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>θ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span>&nbsp;(1/a)</p><p>For inclination&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span> ,t₁ <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>a</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span> (1/a₁) and t₂<span title="Click to copy mathml"><math><mi>α</mi><mi></mi></math></span> (1/a₂)</p><p>Therefore t₂&lt; t₁</p><p>So the sphere will take a longer time to reach plane with smaller inclination.</p>

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