Question

A cylindrical drum has a radius of 0.45m intially at the rest.It is given an angular acceleration of 0.40rad/s2 l.At time t= 8.0s calculate 1)the angular speed of the drum 2)the centripetal acceleration of a point on the rim of the drum 3)the tengential acceleration at the point and 4)the resultant acceleration at the point

Answer 1

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Answer 2

Answer: The angular speed is 3.2 rad/s, the centripetal acceleration is $a_{c}=4.61\ m/s^2$, the tangential acceleration is $a_{t}= 0.18\ m/s^2$ and the resultant acceleration is $a = 4.61\ m/s^2$

Explanation:

Given that,

Radius r = 0.45 m

Angular acceleration $\alpha = 0.40\ rad/s^2$

Time t = 8.0 s

(I). The angular speed of the drum is

$\omega = \alpha\times t$

$\omega = 0.40\times 8.0$

$\omega = 3.2\ rad/s$

The angular speed is 3.2 rad/s.

(II) The centripetal acceleration of a point on the rim of the drum

$a_{c}= r\times\omega^2$

$a_{c}= 0.45\times3.2\times3.2$

$a_{c}=4.61\ m/s^2$

The centripetal acceleration of a point on the rim of the drum is $4.61\ m/s^2$

(III). The tangential acceleration at the point

$a_{t}= r\alpha$

$a_{t}=0.45\times0.40$

$a_{t}= 0.18\ m/s^2$

The tangential acceleration at the point is $0.18\ m/s^2$.

(IV). The resultant acceleration at the point

$a = \sqrt{a_{c}^{2}+a_{t}^{2}}$

$a = \sqrt{(4.61)^2+(0.18)^2}$

$a = 4.61\ m/s^2$

The resultant acceleration at the point is $4.61\ m/s^2$

Hence, this is the required solution.