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3

It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ...


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When the wheel comes in contact with the belt friction will act as there will be relative motion between the belt and the point of contact. Now friction will tend to act on the belt opposite to the velocity of the belt until slipping ceases. To find the work done by the external agency lets consider the energy changes : 1) The K.E of the wheel increases. 2) ...


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If a body of mass m hanged on a string is moving, let uniformly, on a circle fixed relatively to the ground, then an observer G on the ground uses the 2nd Newton Law : $$ \mathbf{F}=m\cdot \mathbf{a} \tag{01} $$ and finds the relation between the force $\mathbf{F}$ and the acceleration $\mathbf{a}$. For observer G there exists a "real" force, the tension ...


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If what you mean is Center of Mass, then, no, it will not change the angular velocity (if there are no other external torque acting on the object, and the object is not attached in any way to other objects, i.e, glued, or attached to a fixed rotating rod or another object, who's rotating axis does not intersect the Center of Mass of the object attached). In ...


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The force of friction acts both towards the centre of the circle and opposite the velocity vector of the car. Strictly speaking, the diagram you have does not show all forces acting on the car but it is enough for purposes of explaining the circular motion. As the text also explains, circular motion always requires a force pointed radially inwards because ...


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Note that you have to swing the pail with a certain minimum speed for the water to stay in. That minimum speed is such that when the pail is at the top of the arc, the rope accelerates the pail downward faster than gravity accelerates the water downward. Otherwise, the water falls out.


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REVISED ANSWER : The tension in the string will become zero when $v^2/r = gsin\theta$. v is still +ve when this happens. v will only become zero later. So you are correct : first T becomes zero (which cannot happen until $\theta > 0$), then the stone moves as a projectile on a parabolic path, then v beomes zero.


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The first green part is the Rodrigue's rotation formula. The second green part is a small angle approximation for $\delta \theta$.



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