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A disk of radius 2.83 m rotates about its axis. Points on the disk's rim undergo tangential acceleration of magnitude 2.53 m/s2. At a particular time the rim has a tangential speed of 1.53 m/s. At a time 0.817 seconds later, what is the tangential speed, v, of a point on the rim, the magnitude of the point's radial acceleration, ar, and the magnitude of its total acceleration, atot?

find:

V=

a(radial)=

a(total)=

Respuesta :

Answer:

Explanation:

v = u + at

= 1.53 + 2.53 x0.817

= 3.6 m/s

Radial acceleration = v² / r

= 3.6 x 3.6 / 2.83

a_r = 4.58 m / s²

a_tot = √ 4.58² + 2.53²

= √ (20.97 + 6.4)

= 5.23 m /s²

Answer:

(a). The tangential speed is 3.597 m/s.

(b). The magnitude of the point's radial acceleration is 4.57 m/s².

(c). The magnitude of its total acceleration is 5.22 m/s².

Explanation:

Given that,

Radius of disk = 2.83 m

Acceleration = 2.53 m/s²

Tangential speed = 1.53 m/s

Time = 0.817 s

(a). We need to calculate the tangential speed

Using equation of motion

[tex]v = u+at[/tex]

Where, v = final velocity

a = acceleration

T = time

Put the value into the formula

[tex]v=1.53+2.53\times0.817[/tex]

[tex]v=3.597\ m/s[/tex]

(b). We need to calculate the magnitude of the point's radial acceleration

Using formula of radial acceleration

[tex]a_{r}=\dfrac{v^2}{r}[/tex]

Put the value into the formula

[tex]a_{r}=\dfrac{3.597^2}{2.83}[/tex]

[tex]a_{r}=4.57\ m/s^2[/tex]

(c). We need to calculate the magnitude of its total acceleration

[tex]a_{tot}=\sqrt{a_{r}^2+a_{t}^2}[/tex]

[tex]a_{tot}=\sqrt{(4.57)^2+(2.53)^2}[/tex]

[tex]a_{tot}=5.22\ m/s^2[/tex]

Hence, (a). The tangential speed is 3.597 m/s.

(b). The magnitude of the point's radial acceleration is 4.57 m/s².

(c). The magnitude of its total acceleration is 5.22 m/s².

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