The scalar product is also known as
a) Additional product
b) Negetive product
c) Dot product
d) None of the option

Answer

A, C

Explanation

A uniform wheel of radius R is set in to rotation about its axis (case-I) at an angular speed `omega`. This rotating wheel is now placed on rough horizontal. Because of its friction at contact, the wheel accelerates forward and its rotation decelerates. As the rotation, if we consider the net momentum before pure rolling and after poure rolling remains constant pure. Before roling the wheel was only rotaing around its axis.
Therefore, angular momentum
`=I_cmomega+m(VxxR)`
`=1/2mR^2(V/R)+mVR`
`=3/2mVR`
Angular momentum
`mVR=1/2mR^2omega=3/2mVR`
`rarr V=(omegaR)/3`