p11_016.pdf

16. The wheel starts turning from rest ( ω 0 =0)at t = 0, and accelerates uniformly at α> 0, which makes

our choice for positive sense of rotation. At t 1 its angular velocity is ω 1 = +10 rev/s, and at t 2 its

angular velocity is ω 2 = +15 rev/s. Between t 1 and t 2 it turns through ∆ θ =60rev,where t 2 −

t 1 =∆ t .

(a) We ﬁnd α using Eq. 11-14:

ω 2 = ω 1 +2 α ∆ θ =

⇒

α = 15 2

10 2

2(60)

−

which yields α =1 . 04 rev/s 2 which we round oﬀ to 1 . 0rev/s 2 .

(b) We ﬁnd ∆ t using Eq. 11-15:

∆ θ = 1

2 ( ω 1 + ω 2 )∆ t =

⇒

∆ t = 2(60)

10+15 =4 . 8s .

ω 1 = ω 0 + αt 1 =

⇒

t 1 = 10

1 . 04 =9 . 6s .

(d) Any equation in Table 11-1 involving θ canbeusedtoﬁnd θ 1 (the angular displacement during

≤

t 1 ); we select Eq. 11-14.

ω 1 = ω 0 +2 αθ 1 =

⇒

θ 1 =

10 2

2(1 . 04) =48rev .

≤