IIT JEE , AIEEE Forum - Physics , Mechanics

KRITKA

PLZ. explain me why is the point of contact of a wheel 'O' ,that rolls without slipping, with a surface at instantaneous rest.

why is the topmost point 'A' moving with velocity twice that of centre of mass 'C' of wheel that rolls with accelaration='a'm/s2?what is its accelaration at time='t's?

what is accelaration of point of contact 'O' at time='t's?

PLZ. give general solution for these.

Thank you!

catch_arnnie

the point of contact is not at instantaneous rest but its relatively at rest with respect to the surface on which the wheel is moving...

moreover, rolling without slipping is pure rolling

the topmost point has 2 motion- one translatory & other rotatry

in translatory motion, the acceleration must be 'a' coz' center of mass has acceration 'a' & since, the body is executing a pure rolling motion, so, it angular acceleration is

(say)& it's tangential acceleration will be R

where R is radius of wheel, which is equal to 'a' i.e. in pure rolling a=R

therefore net acceleration of the topmost point is a + R

= 2a

for a constant force, the acceleration will always remain constant(of any point of the wheel)..

plz rate if you find this useful....

anand_raj

rotatory or translatory motion depends upon frame of refrence

suppose a disk is rolling with speed v . about the centre the motion is rotatory , about the fixd point at the earth the motion is both translatory and rotatory

elessar_iitkgp

Pure rolling can be studied as a combination of translational motion and rotational motion about CM.

In the translational part:
The velocity of each point of the wheel is same as the CM. So the lowermost point has a speed v rightwards.

In the rotational part,
Each point on the rim has a velocity

r directed along the tangent. So the lowermost point has a velocity

r directed leftward (Assuming clockwise rotation)

So the net velocity of the lowermost point is the sum of the forward velocity v and the backward velocity

r . If v =

r then the net velocity of the point of contact is v -

r = 0.

For the topmost point, the velocities due to translation and rotation are in the forward directions. So they add up to v +

r. If v =

r then topmost point moves with speed 2v.

edison

Speed of Point of contact of a rolling sphere = 0

Velocity of center of mass of a rolling sphere = v =

R

Speed of topmost point of the rolling sphere = 2 v

Explantion: To illustrate this le us assume the following example

Consider a freely rotating circular wheel about an axis passing through the center and perpendicular to it, also assume the direction of rotation as CLOCKWISE.

here there is no translation of the wheel as a whole i.e. velocity of center of mass of teh wheel = 0

But speed of any point on the circumference of wheel = v =

R directed tangentially (at any instance).

Thus, the bottom most point has got instantaneous velocity =

R directed tangentially but LEFT to the wheel

Also, the top most point has instantaneous velocity = V_cm directed tangentially but RIGHT to the wheel.

Speed of center of mass = 0

Now if the same wheel is translating in teh forward direction then

Velocity of center of mass = V_cm =

R in the forward direction

Velocity of bottom most / point of contact = V_{cm^{(directed right) +}}

R (LEFT) = 0

Similarly velocity of the topmost point = V_{cm^{(directed right) +}}

R (RIGHT) = 2

R

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