Foda

Chapter 11 Solutions

11.1

( a ) Ktrans =

1 1 mv2 = (10.0 kg)(10.0 m/s) 2 = 500 J 2 2

(b) Krot =

1 1 1 v2 1 Iω2 =  mv 2   2  = (10.0 kg)(10.0 m/s) 2 = 250 J 2 2 2 r  4 

(c) 11.2 K=

Ktotal = Ktrans + Krot = 750 J 1 1 Iω 2 + mv2 2 2 1 4.00 m/s 2 1 (1.60 × 10–2 kg · m2) + (4.00 kg)(4.00 m/s) 2 2  0.100 m  2

K=

= 12.8 + 32.0 = 44.8 J 11.3 W = Kf – Ki = (Ktrans +Krot)f – (Ktrans + Krot)i W= 1 1 1 1 2 v 2 Mv2 + Iω2 – 0 – 0 = Mv2 +  MR 2   2 2 2 2 5  R W = (7/10)Mv2 1 1 1 I v mv2 + Iω2 =  m + 2 v2 where ω = since no slipping. 2 2 2 R  R

or

11.4

K=

Also, Ui = mgh, Uf = 0, and vi = 0 Therefore, 1 I m + 2 v2 = mgh 2 R  2g h [1 + (I/mR 2 )] 1 mR2, so 2 or v disk = 2gh 2 4g h 3 or v ring = gh

Thus,

v2 =

For a disk, I =

v2=

2g h [1 + (1/2)]

For a ring, I = mR2 so v2 = Since vdisk > vring, the disk

reaches the bottom first.

© 2000 by Harcourt College Publishers. All rights reserved.

2

Chapter 11 Solutions ( a ) τ = Iα mg R sin θ = (ICM + mR2)α a= mg R2 sin θ ICM + mR 2 mg R2 sin θ 1 = g sin θ 2mR 2 2 R2
R f

11.5

ahoop =

n

mg

θ

mg sin θ 2 = g sin θ 3 3 mR 2 2 4 The disk moveswith the acceleration of the hoop. 3 adisk = (b) Rf = Iα f = µn = µmg cos θ f Iα/R = = mg cos θ mg cos θ

µ=

 2 g sin θ  1 mR 2 3  2 
R2mg cos θ

=

1 tan θ 3

Goal Solution G: The acceleration of the disk will depend on the angle of the incline. In fact, it should be proportional to g sin θ since the disk should not accelerate when the incline angle is zero, and since a = g whenthe angle is 90° and the disk can fall freely. The acceleration of the disk should also be greater than a hoop since the mass of the disk is closer to its center, giving it less rotational inertia so that it can roll faster than the hoop. The required coefficient of friction is difficult to predict, but is probably between 0 and 1 since this is a typical range for µ. O: We can find theacceleration by applying Newton’s second law and considering both the linear and rotational motion. A free-body diagram will greatly assist in defining our variables and seeing how the forces are related. ∑Fx = mg sin θ – f = maCM ∑Fy = n – mg cos θ = 0 (2) (1)

A:

τ = fr = ICMα =

ICMaCM r

(3)

© 2000 by Harcourt College Publishers. All rights reserved.

Chapter 11 Solutions

3

(a)

21 For a disk, (ICM)disk = mr2. From (3) we find f = 2
Substituting this into (1) gives mg sin θ – 1 maCM 2 so that (aCM)disk =

 1 mr 2 a C M 
r2

=

1 ma CM . 2

2 g sin θ 3

For a hoop, (ICM)hoop = mr2 From (3), f = mr2aCM = maCM r2

Substituting this into (1) gives mg sin θ – maCM so (aCM)hoop = 1 g sin θ 2

2 aCM disk 3 g sin θ 4 Therefore, = = a CM hoop 1 g sin θ 3 2(b)

From (2) we find n = mg cos θ, and f = µn = µmg cos θ Likewise, from equation (1), f = mg sin θ – maCM Setting these two equations equal, µmg cos θ = mg sin θ – 2 mg sin θ 3

so

1 sin θ 1 µ=  = tan θ 3 cos θ 3

L: As expected, the acceleration of the disk is proportional to g sin θ and is slightly greater than the acceleration of the hoop. The coefficient of friction result issimilar to the result found for a block on an incline plane, where µ = tan θ (see Example 5.12). However, µ is not always between 0 and 1 as predicted. For angles greater than 72° the coefficient of friction must be larger than 1. For angles greater than 80° it must be extremely large to make the disk roll without slipping.

© 2000 by Harcourt College Publishers. All rights reserved.

4Chapter 11 Solutions

11.6

3R 2 2 25MR2 1 1 2 2 2 I = M(R1 + R2 ) = M   + R2 = 2 2 32  4  

2

Energy is conserved between x = 2.00 m and x + ∆x. 1 1 2 2 Mv i + Iω i = Mgyf 2 2 1 1 25 2 2 Mv i + MR2 (vi/R2) 2 = Mg ∆x sin θ 2 2 32 57 2 v = g ∆x sin θ 64 i ∆x = 57(2.80 m/s)2 = 1.19 m 64(9.80 m/s2)(sin 36.9°)

So the final position is 2.00 m + 1.19 m = 3.19 m 11.7 1 – ∆x 3.00 m v = = =...