(1)wheel A of radius rA = 14.1 cm is coupled by belt B to wheel C of radius rC = 21.7 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.26 rad/s2. Find the time needed for wheel C to reach an angular speed of 119 rev/min, assuming the belt does not slip. (Hint: If the belt does not slip, the linear speeds at the two rims must be equal.)
（2)The uniform solid block in the figure has mass 25.7 kg and edge lengths a = 0.695 m, b = 1.80 m, and c = 0.130 m. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.
(3)block 1 has mass m1 = 480 g, block 2 has mass m2 = 540 g, and the pulley is on a frictionless horizontal axle and has radius R = 4.6 cm. When released from rest, block 2 falls 71 cm in 5.1 s without the cord slipping on the pulley. (a) What is the magnitude of the acceleration of the blocks? What are (b) tension T2 (the tension force on the block 2) and (c) tension T1 (the tension force on the block 1)? (d) What is the magnitude of the pulley’s angular acceleration? (e) What is its rotational inertia? Caution: Try to avoid rounding off answers along the way to the solution. Use g = 9.81 m/s2.
困難物理題目求解!!!!?
2015-11-11 8:13 am
回答 (1)
2015-11-12 11:41 pm
(1) wheel A of radius Ra = 14.1 cm is coupled by belt B to wheel C of
radius Rc = 21.7 cm. The angular speed of wheel A is increased from
rest at a constant rate of α = 1.26 rad/s2. Find the time needed for
wheel C to reach an angular speed of Nc = 119 rpm, assuming the belt
does not slip.
ωc = 119*2π/60 = 119π/30 = 12.46 rad/s
ωa = 0 + αt = αt = 1.26t
Vb = Ra*ωa = Rc*ωc
ωc = Ra*ωa/Rc
= 14.1*1.26*t/21.7
= 12.46
Answer: t = 12.46*21.7/(14.1*1.26) = 15.221 sec
（2) The uniform solid block in the figure has mass 25.7 kg and edge
lengths a = 0.695 m, b = 1.80 m, and c = 0.130 m. Calculate its
rotational inertia about an axis through one corner and perpendicular
to the large faces.
q = density
m = qabc
I = ∫y^2*dm
= qab∫(0~c)y^2*dy
= qab/3 * y^3
= qab/3 * c^3
= qabc/3 * c^2
= m*c^2/3
= 25.7*0.13^2 / 3
= 0.1448 kg.m^2
(3) m1 = 0.480 kg, m2 = 0.540 g, pulley is on a frictionless horizontal
axle and has radius R = 0.046 m. When released from rest, block 2 falls
0.71 m in 5.1 sec without the cord slipping on the pulley.
(a) a = ?
= 2*L/t^2 ;;; The 3rd Newton's formula
= 2*7.1/5.1^2
= 0.546 m/s^2
(b) What are tension T2 (the tension force on the block 2)
T2 = m2*(g - a)
= 0.54*(9.81 - 0.546)
= 5.003 Newtons
(c) tension T1 (the tension force on the block 1)?
T1 = m1*(g + a)
= 0.48*(9.81 + 0.546)
= 4.97 Newtons
(d) α = ? for the pulley
α = a/R
= 0.546/0.046
= 11.87 rad/s^2
(e) I = ?
m = mass of pully
(T2 - T1)*R = I * α
= (0.5*m*R^2) * (a/R)
= m*a*R/2
m = 2*(T2-T1)/a
I = m*R^2 / 2
= (T2-T1)*R^2 / a
= [m2*(g-a) - m1*(g+a)]R^2 / a
= [(m2-m1)*g - (m2+m1)*a]R^2 / a
= [(m2-m1)g/a - (m1+m2)]R^2
= [(m2 - m1)gt^2 / 2L - (m1 + m2)]R^2
= [0.06*9.81*5.1^2 / 14.2 - 1.02]0.046^2
= 1.230*10^-4 kg.m^2
= 12.30 g.cm^2
radius Rc = 21.7 cm. The angular speed of wheel A is increased from
rest at a constant rate of α = 1.26 rad/s2. Find the time needed for
wheel C to reach an angular speed of Nc = 119 rpm, assuming the belt
does not slip.
ωc = 119*2π/60 = 119π/30 = 12.46 rad/s
ωa = 0 + αt = αt = 1.26t
Vb = Ra*ωa = Rc*ωc
ωc = Ra*ωa/Rc
= 14.1*1.26*t/21.7
= 12.46
Answer: t = 12.46*21.7/(14.1*1.26) = 15.221 sec
（2) The uniform solid block in the figure has mass 25.7 kg and edge
lengths a = 0.695 m, b = 1.80 m, and c = 0.130 m. Calculate its
rotational inertia about an axis through one corner and perpendicular
to the large faces.
q = density
m = qabc
I = ∫y^2*dm
= qab∫(0~c)y^2*dy
= qab/3 * y^3
= qab/3 * c^3
= qabc/3 * c^2
= m*c^2/3
= 25.7*0.13^2 / 3
= 0.1448 kg.m^2
(3) m1 = 0.480 kg, m2 = 0.540 g, pulley is on a frictionless horizontal
axle and has radius R = 0.046 m. When released from rest, block 2 falls
0.71 m in 5.1 sec without the cord slipping on the pulley.
(a) a = ?
= 2*L/t^2 ;;; The 3rd Newton's formula
= 2*7.1/5.1^2
= 0.546 m/s^2
(b) What are tension T2 (the tension force on the block 2)
T2 = m2*(g - a)
= 0.54*(9.81 - 0.546)
= 5.003 Newtons
(c) tension T1 (the tension force on the block 1)?
T1 = m1*(g + a)
= 0.48*(9.81 + 0.546)
= 4.97 Newtons
(d) α = ? for the pulley
α = a/R
= 0.546/0.046
= 11.87 rad/s^2
(e) I = ?
m = mass of pully
(T2 - T1)*R = I * α
= (0.5*m*R^2) * (a/R)
= m*a*R/2
m = 2*(T2-T1)/a
I = m*R^2 / 2
= (T2-T1)*R^2 / a
= [m2*(g-a) - m1*(g+a)]R^2 / a
= [(m2-m1)*g - (m2+m1)*a]R^2 / a
= [(m2-m1)g/a - (m1+m2)]R^2
= [(m2 - m1)gt^2 / 2L - (m1 + m2)]R^2
= [0.06*9.81*5.1^2 / 14.2 - 1.02]0.046^2
= 1.230*10^-4 kg.m^2
= 12.30 g.cm^2
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