The angular acceleration of the disk is defined by  = (3t 2 + 12) rad/s 2 , where t is in seconds. If the disk is originally rotating at 0 = 12 rad/s, determine the magnitude of the velocity and the n and t components of acceleration of point A on the disk when t = 2 s.

Answer :

Answer:

Velocity = 22 m/s

Acceleration = 968 m/s^2

Tangential component of acceleration = 12 m/s^2

Explanation:

The kinematic expression is:

[tex]dw=\alpha dt\\dw=(3t^{2} +12)dt[/tex]

integrating the expression when w =12 rad/s and t=0, we have:

[tex]\int\limits^w_y {dw} \,=\int\limits^2_0 {3t^{2}+12 } \, dt\\ w-12=t^{3} +12t|2-0\\w-12=2^{3} +12*2\\w=44rad/s[/tex]

The velocity at point A is:

[tex]v_{A} =wr_{A} =44*0.5=22m/s[/tex]

α at t=2s is:

[tex]\alpha =3t^{2} +12=3(2^{2} )+12=24rad/s^{2}[/tex]

The normal component of acceleration at point A is equal to:

[tex]a_{A} =w^{2} r_{A} =44^{2} *0.5=968m/s^{2}[/tex]

The tangential component of acceleration at point A is:

[tex]\alpha _{A} =\alpha r_{A} =24*0.5=12m/s^{2}[/tex]

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