The height of a solid cylinder is four times that of its radius. It is kept vertically at time \(t=0\) on a belt which is moving in the horizontal direction with a velocity \(\mathrm{v}=2.45 \mathrm{t}^{2}\) where \(\mathrm{v}\) in \(\mathrm{m} / \mathrm{s}\) and \(t\) is in second. If the cylinder does not slip, it will topple over a time \(t=\) \(\\{\mathrm{A}\\} 1\) second \(\\{\mathrm{B}\\} 2\) second \\{C \(\\}\) \\} second \(\\{\mathrm{D}\\} 4\) second

Short Answer

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Based on the given step-by-step solution, the short answer is: The cylinder will topple over at time \(t=\\{\mathrm{A}\\} 1\) second.

Step by step solution

01

Determine the critical condition for toppling

At the time of toppling, the normal force acting on the base of the cylinder will pass through the point of contact with the belt on its edge. Let the height of the cylinder be \(4r\), and the radius of the base be \(r\). In order to topple, the distance from the center of mass to the point of contact should be equal to the radius 'r'. Using the Pythagorean theorem, we get the critical condition for toppling: \(r^2 = (2r)^2 - (4r)^2\)
02

Simplify the equation

The critical condition equation from the previous step can be simplified as follows: \(r^2 = 4r^2 - 16r^2\) Solving for \(r^2\), we get: \(r^2 = 12r^2\)
03

Calculate the linear acceleration at the time of toppling

Linear acceleration is the horizontal force that causes the cylinder to topple. We can find the linear acceleration (a) at the time of toppling using the formula: \(a = \frac{v^2}{12r}\) Substituting the given velocity formula, we get: \(a = \frac{(2.45t^2)^2}{12r}\)
04

Calculate the angular acceleration at the time of toppling

We know that the cylinder does not slip, so the angular acceleration (α) can be found using the formula: \(a = αr\) Where α is the angular acceleration at the time of toppling.
05

Relate the linear and angular accelerations

Combining the previous steps, we get the equation relating linear and angular accelerations: \(\frac{(2.45t^2)^2}{12r} = αr\)
06

Solve for time 't'

We can now solve for the time 't', at which the cylinder will topple over: \(\frac{(2.45t^2)^2}{12r} = αr\) \((2.45t^2)^2 = 12αr^2\) \((2.45t^2)^2 = 12(αr)r\) \((2.45t^2)^2 = 12(a)r\) \((2.45t^2)^2 = 12\left(\frac{(2.45t^2)^2}{12r}\right)r\) Solving for 't': \(t^4 = 1\) \(t = \pm 1\) The negative value of 't' can be discarded since it has no physical significance in this context. Therefore, the cylinder will topple over at time \(t=\\{\mathrm{A}\\} 1\) second.

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