Problem #1
A bowling ball of Radius R is given an initial velocity of v0 down the lane and a forward spin of ω0 = 3v0/R. It first slips when it makes contact with the lane, but will eventually start to roll without slipping. The coefficient of kinetic friction is μk.
(b) For how long does the ball slide before it begins to roll without slipping?
(c) What is the speed of the bowling ball when it begins to roll without slipping?
(d) What distance does the ball slide down the lane before it starts rolling without slipping?
Answer:
The bowling ball is said to be slipping. That phrase means that the angular acceleration of the bowling all about its CM is NOT related to its linear acceleration, i.e. by a ¹ rα. It also means that we are dealing with kinetic friction. The only point left to resolve is in which direction it points. To do this examine the tangential velocity of the outside rim of the bowling ball, v = Rω0 = 3v0. This means the rim is spinning much faster than the CM is moving forward; friction will act to slow the rim as the rim rubs on the surface of the bowling lane. Friction will point in the direction of the initial velocity. Since friction causes the only torque, the angular acceleration is backwards.
review yoyo translation motion on the x axis:
Σ Fx = max
fk = ma
review cylinder translation motion on the x axis:
ΣFy = may
N − mg = 0
review cylinder rotation motion:
Στcm = Icmαcm
Rfk = Iα
We also know that fk = μN. Since the second equation gives N = mg, we have fk = μmg. The first quation thus yields
a = μg (1)
A bowling ball is a solid sphere and using a table of moments of inertia we find I = (2/5)mR2. With our expression for fk, our third equation becomes
Rμmg = (2/5)mR2αcm
or on rearranging,
αcm = (5/2)μg/R
Time is a kinematics variable. We have initial velocities and accelerations so we can write expressions for
the linear and rotational velocity as a function of time
v(t) = v0 + at = v0 + μgt (3)
and
ω(t) = ω0 – αt = [3v0 – (5/2)μgt]/R
We want the time when
v0 + μgt = 3v0 – (5/2)μgt
Solving for t yields
t = 4v0/7μg
This is the time that is takes for the bowling ball to start to roll without slipping. Plugging this result back into equation (3) gives us the linear velocity of the bowling ball at this and all later times
v(t) = v0 + g[4v0/7μg] = (11/7)v0
Making use of another kinematics formula, we find the distance traveled,
Δx = v0t + ½at2 = v0[4v0/7μg] + ½[μg][4v0/7μg]2
Δx = 36(v0)2/49μg
Problem #2
A solid sphere is sliding (not rolling!) across a frictionless surface with speed v0. It slides onto a surface where the coefficient of kinetic friction is µ. Eventually it will start to roll without slipping.
(a) What is the direction of the frictional force? Explain.
(b) For how long does the sphere slide before it begins to roll without slipping?
(c) What is the speed of the sphere when it begins to roll without slipping?
(d) What distance does the sphere slide it starts rolling without slipping?
The ball is said to be slipping. That phrase means that the angular acceleration of the bowling ball about its CM is NOT related to its linear acceleration, i.e. by a ¹ rα. It also means that we are dealing with kinetic friction. The only point left to resolve is in which direction it points. To do this examine the tangential velocity of the outside rim of the bowling ball, v = 0 since it wasn't rotating. This means the rim will rub on the rough surface as it moves to the right and thus it will experience kinetic friction to the left opposite to the direction of the initial velocity. Since friction causes the only torque, the angular acceleration is forward.
review yoyo translation motion on the x axis:
Σ Fx = max
fk = ma
review cylinder translation motion on the x axis:
ΣFy = may
N − mg = 0
review cylinder rotation motion:
Στcm = Icmαcm
Rfk = Iα
We also know that fk = μN. Since the second equation gives N = mg, we have fk = μmg. The first quation thus yields
a = μg (1)
A bowling ball is a solid sphere and using a table of moments of inertia we find I = (2/5)mR2. With our expression for fk, our third equation becomes
Rμmg = (2/5)mR2αcm
or on rearranging,
αcm = (5/2)μg/R
Our results indicate that the forward motion slows as the forward rotation increases.
Time is a kinematics variable. We have initial velocities and accelerations so we can write expressions for
the linear and rotational velocity as a function of time
v(t) = v0 + at = v0 – μgt (3)
and
ω(t) = ω0 + αt = (5/2)μgt/R (4)
We want the time when |v(t)| = |Rω(t)|, where the absolute bars are there to stress that we are relating magnitudes and must be careful with signs.
Substituting in equation (3) and (4), we get
|v0 – μgt| = |(5/2)μgt|
Solving for t yields
t = 2v0/7μg
This is the time that is takes for the bowling ball to start to roll without slipping. Plugging this result back into equation (3) gives us the linear velocity of the bowling ball at this and all later times
v(t) = v0 – μg[2v0/7μg] = (5/7)v0
Making use of another kinematics formula, we find the distance traveled,
Δx = v0t + ½at2 = v0[2v0/7μg] + ½[μg][2v0/7μg]2
Δx = 12(v0)2/49μg
Problem#3
A ball is placed on an incline as shown in the diagram below. The upper part of the incline is frictionless, so the ball slides but does NOT rotate. At point A, when its speed is 4.50 m/s, it reaches a rough portion of the incline where µk = 0.20. Here the ball starts to slip.
(b) How far down the incline from point A does this occur?
(c) What is the speed of the ball when it starts to roll without slipping?
Answer:
The ball starts to slip at A. That phrase means that the angular acceleration of the bowling ball about its CM is NOT related to its linear acceleration, i.e. by a ¹ rα. It also means that we are dealing with kinetic friction. The only point left to resolve is in which direction it points. To do this examine the tangential velocity of the outside rim of the bowling ball; v = 0 since it wasn't rotating. This means the rim will rub on the rough surface as it moves down the incline and thus it will experience kinetic friction up the incline opposite to the direction of the velocity. Since friction causes the only torque, the angular acceleration is counterclockwise.
review yoyo translation motion on the x axis:
Σ Fx = max
fk + mg sin θ = ma
review cylinder translation motion on the x axis:
ΣFy = may
N – mg cos θ = 0
review cylinder rotation motion:
Στcm = Icmαcm
Rfk = Iα
We also know that fk = μN. Since the second equation gives N = mg cos θ, we have fk = μmg cos θ. The first quation thus yields
a = g(sin θ – μ cos θ) (1)
A bowling ball is a solid sphere and using a table of moments of inertia we find I = (2/3)mR2. With our expression for fk, our third equation becomes
Rμmg = (2/3)mR2αcm
or on rearranging,
αcm = (3/2)μg cos θ/R
Our results indicate that the forward motion slows as the forward rotation increases.
Time is a kinematics variable. We have initial velocities and accelerations so we can write expressions for
the linear and rotational velocity as a function of time
v(t) = v0 + at = v0 + g(sin θ – μ cos θ)t (3)
and
ω(t) = ω0 + αt = (3/2)μgtcos θ/R
We want the time when v(t) = Rω(t), where the absolute bars are there to stress that we are relating magnitudes and must be careful with signs.
Substituting in equation (3) and (4), we get
v0 + g(sin θ – μ cos θ)t = (3/2)μgtcosθ
Collecting the terms involving t together
v0 = g[(3/2)μcosθ – (sinθ – μcosθ)]t
or, more simply
v0 = g[(5/2)μcosθ – sinθ]t ,
Solving for t yields
t = v0/g[(5/2)μ cos θ – sin θ]
known: v0 = 4.50 m/s, θ = 25.00, then
t = 4.50/9.8[(5/2) x 0.20 cos 25.0 – sin 25.0]
t = 4.50/(9.8 x 0.0305) = 15.055 s
Putting this result back into equation (3) gives us the linear velocity of the bowling ball at this and all later times
v(t) = v0 + g(sinθ – μcosθ){v0/g[(5/2)μcosθ – sinθ]}
v(t) = v0{3μ cos θ/[5μcosθ – 2sinθ]}
= 4.50{3 x 0.20 cos 25.0/[5 x 0.20 cos 250 – 2 sin 250]}
v(t) = 40.01 m/s
Making use of another kinematics formula, we find the distance traveled,
Δx = ½(v0 + vf)t
Δx = ½v0{1 + 3μcosθ/[5μcosθ – 2sinθ]}{v0/ g[(5/2)μcosθ – sinθ]}
or, after rearranging,
Δx = [(v0)2/g][8μcosθ – 2 sinθ]/[5μcos θ – 2 sinθ]2
Then,
Δx = [(4.50)2/9.80][8 x 0.20 cos 25.00 – 2 sin 25.00]/[5 x 0.20 cos 25.00 – 2 sin 25.00]2
Δx = (2.066)(0.605)/(3.729 x 10-3)
Δx = 335.13 m
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