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In the sae figure. If v and omega both a...

In the sae figure. If `v` and `omega` both are constant, then find linear acceleration of point M,N,R and S in terms of `R,omega,hati` and `hatj` where R is the radius disc.

Text Solution

Verified by Experts

There is only one acceleration
`a_(r)=Romega^(2)` (towards centre)
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Knowledge Check

  • A ring of radius R rolls on a horizontal surface with constant acceleration a of the centre of mass as shown in figure. If omega is the instantaneous angular velocity of the ring. Then the net acceleration of the point of contact of the ring with gound is

    A
    zero
    B
    `omega^(2)R`
    C
    `a`
    D
    `sqrt(a^(2)+(omega^(2)R)^(2))`

  • A disc of radius R rolls on a horizontal ground with linear acceleration a and angular acceleration alpha as shown in Fig. The magnitude of acceleration of point P as shown in the figure at an instant when its linear velocity is v and angular velocity is omega will be a

    A
    `sqrt((a+ralpha)^(2)+(romega^(2))^(2))`
    B
    `(ar)/R`
    C
    `sqrt(r^(2)alpha^(2)+r^(2)omega^(4))`
    D
    `ralpha`

  • The linear velocity of a rotating body is given by v =omega xx r , where omega is the angular velocity and r is the radius vector. The angular velocity of a body omega=hati-2hatj+2hatk and their radius vector r=4hatj-3hatk, |v| is -

    A
    `sqrt(29)` units
    B
    31 units
    C
    `sqrt(37)` units
    D
    `sqrt(41)` units

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