Bowling Alley Physics

Question

You slide a bowling ball down the lane. The bowling ball (which youcan assume is spherical and homogeneous) has a mass of M =6.30 kg and a radius of R = 0.21 m. The speed of thecenter of mass of the ball is initially v0 =8.70 m/s. Just as you release the ball, its initital angularvelocity is 0 rad/s (it is not rotating initially). As the ballslips along the floor, friction causes it to start rotating. As itdoes, it is eventually rolling without slipping. If the coefficientof kinetic friction between the ball and the floor isk = 0.16, how far does the bowling ballslip before it starts rolling (without slipping)?

x =
Hint: To start off, do your FBD like usual(weight, normal
force, kinetic friction). Assign positive directions. Sum
up your forces and torques. You should only have a
kinetic friction force. You should have a torque due to
kinetic friction as well (it's not acting on the COM).

Solve for alpha in the torques and acceleration in the
forces. To find t (which you must to start), integrate
both of those (don't forget to add omega initial and
velocity initial). Make these two integrals equal to each
other by using Vcom = w*r(rolling)
You slide a bowling ball down the lane. The bowling ball (which youcan assume is spherical and homogeneous) has a mass of M =6.30 kg and a radius of R = 0.21 m. The speed of thecenter of mass of the ball is initially v0 =8.70 m/s. Just as you release the ball, its initital angularvelocity is 0 rad/s (it is not rotating initially). As the ballslips along the floor, friction causes it to start rotating. As itdoes, it is eventually rolling without slipping. If the coefficientof kinetic friction between the ball and the floor isk = 0.16, how far does the bowling ballslip before it starts rolling (without slipping)?

x =
Hint: To start off, do your FBD like usual(weight, normal
force, kinetic friction). Assign positive directions. Sum
up your forces and torques. You should only have a
kinetic friction force. You should have a torque due to
kinetic friction as well (it's not acting on the COM).

Solve for alpha in the torques and acceleration in the
forces. To find t (which you must to start), integrate
both of those (don't forget to add omega initial and
velocity initial). Make these two integrals equal to each
other by using Vcom = w*r(rolling) To start off, do your FBD like usual(weight, normal
force, kinetic friction). Assign positive directions. Sum
up your forces and torques. You should only have a
kinetic friction force. You should have a torque due to
kinetic friction as well (it's not acting on the COM).

Solve for alpha in the torques and acceleration in the
forces. To find t (which you must to start), integrate
both of those (don't forget to add omega initial and
velocity initial). Make these two integrals equal to each
other by using Vcom = w*r(rolling)

Explanation / Answer

The simplest solution is to use conservation of angularmomentum taking angular momentum about the point of contact. M v0 R = M vf R + I = Mvf R + 2/5 M R2 vf / R = 7 / 5 M Rvf Then vf = 5 v0 / 7 2 a s = vf2 -v02 = (25 / 49 - 1)v02 Since F = - M g      a =- g Then s = v02 12 / (49 g) = 11.8m A little more complicated way to do the the problem butsimpler than integrating is vf = v0 + a t = v0 - gt vf = R = tR       where torque = M gR and = M g R / I vf = R = t R = M gR2 t / I = 5 g t / 2 v0 - a t = v0 - g t = 5 gt / 2     and t = 2 v0 / (7 g) s = v0 t - 1/2 a t2 = 2v02 / (7 g) - 1/2 (4 / 49) *(v0 / ( g))2 * g = 12v02 / (49 g) If you don't like either of these solutions you can still getthe same result by using conservation of energy Since F = - M g      a =- g Then s = v02 12 / (49 g) = 11.8m A little more complicated way to do the the problem butsimpler than integrating is vf = v0 + a t = v0 - gt vf = R = tR       where torque = M gR and = M g R / I vf = R = t R = M gR2 t / I = 5 g t / 2 v0 - a t = v0 - g t = 5 gt / 2     and t = 2 v0 / (7 g) s = v0 t - 1/2 a t2 = 2v02 / (7 g) - 1/2 (4 / 49) *(v0 / ( g))2 * g = 12v02 / (49 g) If you don't like either of these solutions you can still getthe same result by using conservation of energy

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