Question

The magnitude of acceleration of a point on the
wheel's rim at t = 1 s, whose radius is 2 m and
is spinning with angular velocity a that varies with
time as a = 10 t2 rad/s, is
(1) 40root26 m/s2
(2) 40 m/s2
(3) Zero
(4) 20 m/s2​

Answer 1

Answer: 1) 40√26 m/s^2

Explanation:

Answer 2

The magnitude of acceleration of a point on the  wheel's rim is $40\sqrt{26}\ m/s^2$

Option (1)

Explanation:

Given that,

Time t = 1

Radius = 2 m

Angular velocity $\omega= a$

Here, $a =10t^2$

We need to calculate the angular acceleration

Using formula of angular acceleration

$\alpha=\dfrac{dw}{dt}$

$\alpha=\dfrac{d}{dt}(10t^2)$

$\alpha=20t$

At t = 1,

$\alpha=20\ m/s^2$

We need to calculate the tangential acceleration

Using formula of tangential acceleration

$a_{t}=\alpha r$

Put the value into the formula

$a_{t}=20\times2$

$a_{t}=40\ m/s^2$

We need to calculate the centripetal acceleration

Using formula of centripetal acceleration

$a_{c}=\dfrac{v^2}{r}$

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

Put the value into the formula

$a_{c}=10^2\times2$

$a_{c}=200\ m/s^2$

We need to calculate the magnitude of acceleration of a point on the  wheel's rim

Using formula of acceleration

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

$a=\sqrt{(40)^2+(200)^2}$

$a=40\sqrt{26}\ m/s^2$

Hence, The magnitude of acceleration of a point on the  wheel's rim is $40\sqrt{26}\ m/s^2$

# Learn more

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