Question

(a) What is the tangential acceleration of a bug on the rim of a 10.0 in. diameter disk if the disk moves from rest to an angular speed of 79 revolutions per minute in 4.0 s? m/s2

(b) When the disk is at its final speed, what is the tangential velocity of the bug? m/s

(c) One second after the bug starts from rest, what is its tangential acceleration? m/s2

What is its centripetal acceleration? m/s2

What is its total acceleration? m/s2 ° (relative to the tangential acceleration)

Answer 1

dadius os tue rim Pus's (012) 2.065 mse2

Answer 2

Thank you very much for this!

Answer 3

SOLUTION :

a.

Angular acceleration 

= Change in angular speed (in rad/sec)  / time in sec.

= (79* 2 pi / 60) / 4 

= 2.0682 rad/sec^2 

So, tangential acceleration 

= angular acceleration  in rad/sec^2 * radius of disk in m 

= 2.0682 * (10/2) * (2.54/100)

= 0.2627 m/sec^2 (ANSWER) .

b.

Tangential velocity 

= Tangential acceleration * time 

= 0.2667 * 4 

= 1.05 m/s (ANSWER).

c.

Angular speed after 1 sec 

= angular acceleration * time

= 2.0682 * 1 

= 2.0682 rad/sec

Tangential speed after 1 sec

= angular speed after 1 sdec * radius

= 2.0682 * (5 * 2.54/100)

= 0.2627 m/s 

Centripetal acceleration 

= v^2 / r

= (0.2627)^2 / (5 * (2.54/100)

= 0.5434 m/sec^2 (ANSWER)

Resultant acceleration (Total acceleration)

= sort ( centripetal acceleration ^2 + tangential acceleration^@)

= sqrt(0.5434^2 + 0.2627^2) (two accelerations are at right angle).

= 0.6036 m/sec^2 (ANSWER).