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March 29, 2015

Posted by **lost** on Wednesday, May 16, 2012 at 1:12am.

find angular velocity just after blow, velocity once roles without slipping, and KE just after hit.

- physics college -
**Elena**, Wednesday, May 16, 2012 at 6:09pm1.

The linear momentum is p =m•vₒ.

An angular momentum is

L = p•(2•R/3) = m•vₒ•(2•R/3).

Moment of inertia of the sphere is

I =2•m•R²/5.

L = I• ωₒ= (2•m•R²/5)•ωₒ.

m•vₒ•(2•R/3) = (2•m•R²/5)•ωₒ.

ωₒ = 5•vₒ/3•R.

Since the force is applied below the center line, the spin is

backward, i.e., the ball will slow down,

ωₒ = - 5•vₒ/3•R.

2.

M = F•R =μ•g•R is the torque of the friction force,

then an angular acceleration is

ε = M/I = μ•g•R/(2•m•R²/5) = 5• μ•g/2•R.

v =vₒ - a•t = vₒ- μ•g•t.

ω = ωₒ+ε•t= -(5•vₒ/3•R) +(5• μ•g/2•R)•t.

Since v =ω•R,

v = - (5•vₒ/3•R) •R +(5• μ•g/2•R) •R•t =

= - (5•vₒ/3)+ (5•μ•g/2) •t,

vₒ - μ•g•t = - (5•vₒ/3) + (5•μ•g/2) •t ,

vₒ + (5•vₒ/3) = μ•g•t + (5•μ•g/2) •t.

16•vₒ = 21•μ•g•t,

t = 16•vₒ/21•μ•g.

v = vₒ- μ•g•t = vₒ- μ•g•(16•vₒ/21•μ•g) =

= 5• 21 =0.238•vₒ.

3.

KE1 = m• vₒ²/2 + I•ωₒ²/2 =

= m• vₒ²/2 + (2•m•R²/5)•( 5•vₒ/3•R)²/2 =

=19• m• vₒ²/18 =1.056• m• vₒ².

KE2 = m• v²/2 + I•ω²/2 =

= m• v²/2 + (2•m•R²/5)•v²/2•R² = 0.7•m•v²=

0.7•m•(0.238•vₒ)²= 0.0397•m•vₒ².

ΔKE = - W(fr) = 1.056• m• vₒ² - 0.0397•m•vₒ² =

=1.016•m•vₒ².

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