Question #7964

Q2:(a) What is the tangential acceleration of a bug on the rim of a 8.0 in. diameter disk if the disk moves from rest to an angular speed of 79 revolutions per minute in 4.0 s? (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? What is its centripetal acceleration? What is its total acceleration? ° (relative to the tangential acceleration)

Expert's answer

(a) First of all, convert 79 rpm to radians per second, which we will call w.

w = 79 rev/min* (2 pi rad/rev) * (1 min/60s) = 8.27 rad/s

Divide that final angular velocity by 4 seconds to obtain the angular acceleration. Call that number "alpha"

alpha = (8.27 rad/s)/4.0s = 2.07 rad/s^2

The tangential acceleration of the bug

while the disk is speeding up is

a_t = alpha*R

where R = 4.5 in/39.37 in/meter = 0.1143 m. Therefore

a_t = 0.236 m/s^2

b) The tangential velocity at full speed is v_t = R*w

c) same tangential acceleration as (a)

centripetal acceleration @ t=1 s is

a_c (@ t=1) = R w^2,

where w is 1/4 of the value calculated in (a), since you are only 1/4 way through the acceleration interval.

For the total acceleration at t=1 s, use

a = sqrt[a_t^2 + a_c^2]

The tangent of the total acceleration vector relative to tangential direction is (a_c)/(a_t).

w = 79 rev/min* (2 pi rad/rev) * (1 min/60s) = 8.27 rad/s

Divide that final angular velocity by 4 seconds to obtain the angular acceleration. Call that number "alpha"

alpha = (8.27 rad/s)/4.0s = 2.07 rad/s^2

The tangential acceleration of the bug

while the disk is speeding up is

a_t = alpha*R

where R = 4.5 in/39.37 in/meter = 0.1143 m. Therefore

a_t = 0.236 m/s^2

b) The tangential velocity at full speed is v_t = R*w

c) same tangential acceleration as (a)

centripetal acceleration @ t=1 s is

a_c (@ t=1) = R w^2,

where w is 1/4 of the value calculated in (a), since you are only 1/4 way through the acceleration interval.

For the total acceleration at t=1 s, use

a = sqrt[a_t^2 + a_c^2]

The tangent of the total acceleration vector relative to tangential direction is (a_c)/(a_t).

## Comments

## Leave a comment