Answer:
a) 1510/[tex]s^{2}[/tex]
b) 52.55[tex]m/s^{2}[/tex]
c) 52.55[tex]m/s^{2}[/tex]
Explanation:
Assuming:
1) Point A is on the periphery of the disk and point B is on periphery of pulley
2) Disk has a radius of 100mm
3) Disk has a radius of 50mm
4) The wheel and the disc experience only rotational motion and there is no slip between he belt and the pulley
a)Since acceleration a of any point on a rotating disc is given as:
a=rα
⇒ α=a/r
⇒ α=[tex]\frac{151 m/s^{2} }{100mm}\\[/tex] = 1510/[tex]s^{2}[/tex]
Since pulley and disc are attached to each oter theay have the same angular acceleration.
b) Since point B is on periphery of the pulley, it experiences only rotational motion and no translational motion. Hence:
a=rα
⇒ a= 50mm × 1051 [tex]s^{-2}[/tex] = 52.55[tex]m/s^{2}[/tex]
c) Since belt and point B on the pulley experience same motion (pulley is run by the belt), the magnitude of acceleration of point C is same as point B i.e. 52.55[tex]m/s^{2}[/tex]