
A solid homogeneous cylinder of height h and base radius r is kept vertically on a conveyor belt moving horizontally with an increasing velocity $v = a + b{t^2}$ . If the cylinder is not allowed to slip, the time when the cylinder is about to topple is given as $\dfrac{{xgr}}{{bh}}$. Find $x$.
Answer
163.2k+ views
Hint: We can always assume that the centre of mass is where the force acts. Any force acting on a rigid body adds to the formula $F=ma$, where $a$ represents the centre of mass's translational acceleration. This holds true regardless of whether the reaction force is parallel to the centre of mass.
Formula Used:
$F = ma$
Net force, $N = mg$
Here, $m$ is the mass of the body and $g$ is the acceleration due to gravity.
Complete step by step solution:
As according to the question it is given that the net force in vertical direction is equal to zero.
$N = mg$
Acceleration will be,
$a = \dfrac{{dv}}{{dt}} = 2bt$
As for the force according to the question,
$F = ma = 2btm$
This force acts on the geometric centre, which corresponds to the centre of mass.
The normal would have been adjusted to the farthest position to provide the most torque when the cylinder was just about to topple. (typical behaviours at the base circle's perimeter). When we write the torque equation around the point at the base of the cylinder, we obtain,
$N(R) = F\left( {\dfrac{h}{2}} \right)$
As we mentioned above the formulas, by using them in the above equation we get,
$mg(R) = 2mbt\left( {\dfrac{h}{2}} \right)$
By cancelling out the quantities which are at both side, we get the equation as,
$gR = bth$
According to the question, we need the value of the time when the cylinder is about to topple,
$t = \dfrac{{gR}}{{bh}}$
Since, in question the value is given as $\dfrac{{xgr}}{{bh}}$ by comparison of both equation we get the value of x as,
$x = 1$
Therefore, the correct answer for the value of x is $1$.
Note: The combined effect (the sum) of all pushing and pulling forces that are really operating on an object is the net force. An object will accelerate in the direction of the net force if the pushing and pulling forces acting on it are not equal (a net force acts).
Formula Used:
$F = ma$
Net force, $N = mg$
Here, $m$ is the mass of the body and $g$ is the acceleration due to gravity.
Complete step by step solution:
As according to the question it is given that the net force in vertical direction is equal to zero.
$N = mg$
Acceleration will be,
$a = \dfrac{{dv}}{{dt}} = 2bt$
As for the force according to the question,
$F = ma = 2btm$
This force acts on the geometric centre, which corresponds to the centre of mass.
The normal would have been adjusted to the farthest position to provide the most torque when the cylinder was just about to topple. (typical behaviours at the base circle's perimeter). When we write the torque equation around the point at the base of the cylinder, we obtain,
$N(R) = F\left( {\dfrac{h}{2}} \right)$
As we mentioned above the formulas, by using them in the above equation we get,
$mg(R) = 2mbt\left( {\dfrac{h}{2}} \right)$
By cancelling out the quantities which are at both side, we get the equation as,
$gR = bth$
According to the question, we need the value of the time when the cylinder is about to topple,
$t = \dfrac{{gR}}{{bh}}$
Since, in question the value is given as $\dfrac{{xgr}}{{bh}}$ by comparison of both equation we get the value of x as,
$x = 1$
Therefore, the correct answer for the value of x is $1$.
Note: The combined effect (the sum) of all pushing and pulling forces that are really operating on an object is the net force. An object will accelerate in the direction of the net force if the pushing and pulling forces acting on it are not equal (a net force acts).
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