Respuesta :
Answer:
a) v_com= Rω
b) -2.254 m/[tex]s^{2}[/tex]
c) 51.2 rad/[tex]s^{2}[/tex]
d) t=1.08 seconds
e) x=7.865m
f) v_roll=6.07m
Explanation:
Initially, the ball is travelling with v_com=v_0
Wen not rotating, at the initial stage the ball must be sliding along the surface.
This motion therefore generates a frictional force F_r at the point of contact.
Let the velocity at the point of contact be v_bottom
v_bottom=v_com-Rω
Therefore when ω=0, v_bottom=v_com
So when the ball begins rolling
v_com= Rω
F_r=μ_rmg
〖-F〗_r=ma_com
a_com=(〖-μ〗_r mg)/m
a_com=-μ_rg
a_com=-(0.23)(9.8)
a_com=-2.254m/s^2
Te negative sow decrearse
[tex]\alpha[/tex]=(μ_r mgR)/I = (〖5μ〗_r mgR)/2mRR
=(〖5μ〗_r g)/2R
=(5*(0.23)*(9.8))/(2*0.11)
=51.2 rad/s^2
t=v_0/(〖-a〗_com+Rα)
=8.5/(2.255+0.11*(51.2))
=8.5/7.886
=1.08 seconds
X=v_0 t+1/2 a_com t^2
X=8.5*(2.254) - 1/2 (2.254)*〖1.08〗^2
=7.865m
v_roll=v_0+a_com t_r
=8.5-(2.254)(1.08)
=6.07m/sec
Initially, the ball is travelling with v_com=v_0
Wen not rotating, at the initial stage the ball must be sliding along the surface.
This motion therefore generates a frictional force F_r at the point of contact.
a) Let the velocity at the point of contact be v_bottom
v_bottom=v_com-Rω
Therefore when ω=0, v_bottom=v_com
So when the ball begins rolling
v_com= Rω
b) F_r=μ_rmg
〖-F〗_r=ma_com
a_com=(〖-μ〗_r mg)/m
a_com=-μ_rg
a_com=-(0.23)(9.8)
a_com=-2.254m/s^2
Te negative sow decrearse
c) α=(μ_r mgR)/I = (〖5μ〗_r mgR)/2mRR
=(〖5μ〗_r g)/2R
=(5*(0.23)*(9.8))/(2*0.11)
=51.2 rad/s^2
d) t=v_0/(〖-a〗_com+Rα)
=8.5/(2.255+0.11*(51.2))
=8.5/7.886
=1.08 seconds
e) X=v_0 t+1/2 a_com t^2
X=8.5*(2.254) - 1/2 (2.254)*〖1.08〗^2
=7.865m
f) v_roll=v_0+a_com t_r
=8.5-(2.254)(1.08)
=6.07m/sec
