The following bodies, (1) a ring (2) a disc (3) a solid cylinder (4) a solid sphere of same mass 'm' and radius 'R' are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ---. [Mark the body as per their respective numbering given in the question]
The equations for an object rolling down an inclined plane without slipping are:
· Force equation: mg sinθ - f_s = ma
· Torque equation: f_s R = Iα
Since a = αR, we can write f_s = Iα/R = Ia/R². Substituting this into the force equation: mg sinθ - Ia/R² = ma mg sinθ = a (m + I/R²) a = (mg sinθ) / (m + I/R²)
The time taken to travel a distance S is given by S = ½ at², which means t ∝ 1/√a. Therefore, the object with the largest acceleration (a) will arrive first.
The problem i
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The equations for an object rolling down an inclined plane without slipping are:
· Force equation: mg sinθ - f_s = ma
· Torque equation: f_s R = Iα
Since a = αR, we can write f_s = Iα/R = Ia/R². Substituting this into the force equation: mg sinθ - Ia/R² = ma mg sinθ = a (m + I/R²) a = (mg sinθ) / (m + I/R²)
The time taken to travel a distance S is given by S = ½ at², which means t ∝ 1/√a. Therefore, the object with the largest acceleration (a) will arrive first.
The problem is analyzed for different bodies:
· Ring: I = mR², a = (g sinθ) / (m + mR²/R²) = (g sinθ) / 2 = 0.50 g sinθ
· Disc / Solid Cylinder: I = ½ mR², a = (g sinθ) / (m + ½mR²/R²) = (2g sinθ) / 3 ≈ 0.67 g sinθ
· Solid Sphere: I = (2/5)mR², a = (g sinθ) / (m + (2/5)mR²/R²) = (5g sinθ) / 7 ≈ 0.71 g sinθ
Conclusion: The solid sphere has the greatest acceleration, so it will come down first.
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<p>The equations for an object rolling down an inclined plane without slipping are:</p><p><!- [if !supportLists]->· <!- [endif]->Force equation: mg sinθ - f_s = ma</p><p><!- [if !supportLists]->· <!- [endif]->Torque equation: f_s R = Iα</p><p>Since a = αR, we can write f_s = Iα/R = Ia/R².<br>Substituting this into the force equation:<br>mg sinθ - Ia/R² = ma<br>mg sinθ = a (m + I/R²) <br>a = (mg sinθ) / (m + I/R²)</p><p>The time taken to travel a distance S is given by S = ½ at², which means t ∝ 1/√a. Therefore, the object with the largest acceleration (a) will arrive first.</p><p>The problem is analyzed for different bodies:</p><p><!- [if !supportLists]->· <!- [endif]-><strong>Ring:</strong> I = mR², a = (g sinθ) / (m + mR²/R²) = (g sinθ) / 2 = 0.50 g sinθ</p><p><!- [if !supportLists]->· <!- [endif]-><strong>Disc / Solid Cylinder:</strong> I = ½ mR², a = (g sinθ) / (m + ½mR²/R²) = (2g sinθ) / 3 ≈ 0.67 g sinθ</p><p><!- [if !supportLists]->· <!- [endif]-><strong>Solid Sphere:</strong> I = (2/5)mR², a = (g sinθ) / (m + (2/5)mR²/R²) = (5g sinθ) / 7 ≈ 0.71 g sinθ</p><p><strong>Conclusion:</strong> The solid sphere has the greatest acceleration, so it will come down first.</p><p><img ></p>
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