# Newton's laws--Block sliding up a frictionless semi-circular track

• sachin
The equation of motion for the small block is:##u## is the velocity of the small block of mass ##m##.## r ## is the radius of the semicircle.## \theta ## is the angle of the mass ##m## w.r.t vertical.## \frac{m}{r} \dot u - Nm \dot r = 0where:##N## is the total mass of the two blocks##mg## is the mass of the big block##m## is the mass of the small blockf
View attachment 312968

W.r.t the ground the velocity of the sled when the mass is at some angle ##\theta## is represented by the green arrow

W.r.t the ground the velocity of the mass on the sled in the horizontal direction is the projection of the red arrow onto the horizontal direction plus the green arrow.

In general, the velocity of the sled in the horizontal direction will not equal the velocity of the mass in the horizontal direction.

It is at point A, when the projection of the red arrow onto the green arrow is ##0##, that the velocity of the sled matches the velocity of the mass.
yes,its all true,the red arrow has only horizontal and vertical components,has not got any component along the common normal i.e perpendicular to the red one,for the bodies to be in contact the velocities along the common normal should be the same,how the contact still remains ?

yes,its all true,the red arrow has only horizontal and vertical components,has not got any component along the common normal i.e perpendicular to the red one,for the bodies to be in contact the velocities along the common normal should be the same,how the contact still remains ?
They are the same. The velocities of both blocks along the common normal is ##0##. The ground is interfering with the sleds ability to translate in the common normal direction. The sled cannot translate in the direction of the orange arrow.

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Lnewqban
They are the same. The velocities of both blocks along the common normal is ##0##. The ground is interfering with the sleds ability to translate in the common normal direction. The sled cannot translate in the direction of the orange arrow.
yes,its quite clear now,thanks a lot.
View attachment 312996

They are the same. The velocities of both blocks along the common normal is ##0##. The ground is interfering with the sleds ability to translate in the common normal direction. The sled cannot translate in the direction of the orange arrow.

View attachment 312996
yes,its all clear,hope the same relationship holds true for displacement and acceleration also,thanks.

Lnewqban, erobz and Dale