New answers tagged

-1

we know : $$F=ma$$ $\therefore \ a\ _{tangent} = dv\ _{tangent}/dt $ $ \int_0^v dv\ _{tangent} = \int_0^t dt \ a _{tangent}$ = If F is constant , |a| is constant . $v\ _{tangent} =a_t \ t$ $a_{c}= v^2_{tan}/r_{inst}$


0

Why didn't he consider the centripetal force as a force acting on that object. Because Centripetal force is not a force acting on the body. The Centripetal force is not necessarily a unique force in its own right. As an answer pointed out, The "centripetal force" is simply defined as the net force acting towards the centre. If there were some more ...


4

The first part of your question is largely philosophical - can any "perfect" shape or motion really exist in the real world? Everything is made of atoms that are ultimately jiggling around and changing position, so before we can say if a perfectly circular stone path can exist in space, we have to define what we really mean by the position of the ...


11

As you suspect the wire cannot be perfectly horizontal nor, if its weight is considered, perfectly straight. At any point in time there are two forces acting on the ball,( ignoring the weight of the rope) : gravity (mg) and the tension due rope . The vector sum of these must be equal to $mv^2/r$ in the radial direction. To counter the vertical gravitational ...


5

The vertical component of the tension in the string balances the force due to gravity. Since this component vanishes if the string is horizontal, the string must sweep out a (possibly very squashed) cone.


-1

First of all let me clear you that whenever a body moves in a circular path there is always an acceleration radially inwards which we refer to as Centripetal acceleration. And then there must be a force associated with this acceleration again radially inward. We call this force as Centripetal force. Here is the proof of the presence of a centripetal ...


1

The horizontal component of the external force coming from the hip of the person rotating the hoop must be mr$ω^2$ where m is the mass of the hoop, ω is the angular velocity (in rad./s), and r is measured from the center of mass of the hoop to the center of the circle in which it is moving. Since the hips are moving and are not perfect circles, r (measured ...


2

An rotating object will deform if it isn't resilient enough to provide the required centripetal force. Take for example the manual skill of spreading out pizza dough by tossing it in the air with a lot of rotation. A wooden disk would keep its shape. The dough does have elasticity (making it tedious to try and spread it out with a roller) but the dough does ...


1

He is doing right . Centripetal force is not any particular force ,any force acting in the radial direction behaves as a centripetal force,so your thinking that we should add mv²/R as centripetal force in LHS always is wrong. The net of all the forces acting in the radial direction is the centripetal force.The role of centripetal force can be played by any ...


0

The two-wheel model shown in the question is good start for determining the loading on the tires. But this would only result with the total load on the front tires, and the total load on the back tires, without any further details on how these loads are distributed left-to-right. To get there, you also need to consider the height of the center of mass ...


0

Try considering this Since the car is driven by one engine let us assume all 4 wheels have same velocities at all time for simplicity. And the weight of car is divided quite uniformly over the 4 wheels as well Sonnow that we have made the m•(v^2) part of the equation same for all the wheels let us move on to the Raius of curvature part And as for Radius os ...


2

Why is the centripetal force non conservative while the centrifugal isn't? As regards to the title of your post, the centripetal or centrifugal force may be conservative or non conservative. For example, if the centripetal force is gravity then it is conservative. The equal but opposite centrifugal force (pseudo force) that exists only in the rotating (non ...


0

What force an object is experiencing depends upon the reference frame of the observer. If you sit in the rock's frame (the one which is rotating), the only force acting on it (excluding frictional forces at contact) is fictitious centrifugal force which directs outwards away from the disk's centre. Generally speaking, fictitious force comes into the picture ...


1

Draw a tangent to the disk at a point. the rock has tendency to move along such a tangent at each instant(due to its inertia). hence it feels as though it is pushed outwards w.r.t. the disk. and for the same reason it needs a force along the radial direction to keep up with the disk.


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