Assume a billiard ball is a solid sphere. Its radius is 3 cm and mass is 500 gra

Question

Assume a billiard ball is a solid sphere. Its radius is 3 cm and mass is 500 grams. The coefficient of static friction between the ball and the surface is .70.

a) Where should you hit the ball so that it experiences pure rolling motion without the need for any friction force? What is its acceleration if the force is 10 Newtons?

h=1.2cm above the center of mass. a=20m/s^2

b) What is the highest you can hit it with 10 Newtons and still have pure rolling motion? What is the acceleration now?

h=2.64cm above the center of mass, a=26.86m/s^2

c) What is the lowest you can hit it with 10 N for pure rolling motion? What is it’s a?

h=0.24cm below the center of mass, a=13.14m/s^2

I need help with the step-by-step.

Explanation / Answer

a)

let the point of impact be at height 'h' from center of mass i.e. center of the sphere.

let 'q' be the angle made by the line joining the center and point of impact with horizontal.

As required condition is there should be no need of frictional force for pure rolling, only external torque is due to the force(10 N).

Fsinq = (2/5)mr2(alpha)   [where alpha is angular acceleration]

10(h/r)(r) = (0.4)(0.5)(0.03)2(alpha) -----------------equation 1

condition for pure rolling: point that is in contact with surface should have zero net acceleration

So (alpha)(r) = a where 'a' is linear accelaration

As net force is 10 newtons (there is no frictional force as per condition given in question).

a= 10/0.5;

=> a=20m/sec2

alpha = 666.67 rad/sec2

Substitute alpha in equation 1 and you get h = 1.199 cm.

b) let the highest hight you can hit be 'h'

This point is the point at which the limiting friction acts in forward direction.

limiting friction is mg(coff) = (0.5)(10)(0.7) = 3.5 N

Torque equations:

Frsinq - (limiting friction)(r) = (2/5)mr2(alpha)

(10)(r)(h/r) - 3.5r = (0.4)(0.5)(0.03)2(alpha) ---------------equation 1

condition for pure rolling: point that is in contact with surface should have zero net acceleration

So (alpha)(r) = a where 'a' is linear accelaration

Force equations:

net force = 10 + 3.5 =13.5N [as limiting friction is acting in forward direction]

=> a =27m/sec2

alpha = a/r = 900rad/sec2

Substitute alpha in equation 1 and you get h (calculaions not done, do it yourself for clear understanding)

c)

let the lowest hight you can hit be 'h'

This point is the point at which the limiting friction acts in backward direction.

limiting friction is mg(coff) = (0.5)(10)(0.7) = 3.5 N

Torque equations:

Frsinq + (limiting friction)(r) = (2/5)mr2(alpha)

(10)(r)(h/r) + 3.5r = (0.4)(0.5)(0.03)2(alpha) ---------------equation 1

condition for pure rolling: point that is in contact with surface should have zero net acceleration

So (alpha)(r) = a where 'a' is linear accelaration

Force equations:

net force = 10 - 3.5 =6.5N [as limiting friction is acting in forward direction]

=> a =13m/sec2

alpha = a/r = 433.33rad/sec2

Substitute alpha in equation 1 and you get h (calculaions not done, do it yourself for clear understanding)

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