What are the full solutions of some mechanics questions:
Q1. Part of a hallow spherical shell, centre O and radius a, is removed to form a bowl with a plane circular rim. The bowl is fixed with the rim uppermost and horizontal. The centre of the circular rim is (4a/3) vertically above the lowest point of the bowl. A marble is projected from the lowest point of the bowl with speed u. Find the minimum value of u for which the marble will leave the bowl and not fall back to it. [Ans: (17ga/3)^(1/2) ]
Q2. A particle P of mass m is attached to two strings AP and BP. The points A and B are on the same horizontal level and AB= (5a/4).
The string AP is inextensible and AP = (3a/4)
The string BP is elastic and BP= a
The modulus of elasticity of BP is k.
Show that the natural length of BP is (5ka)/(3mg+5k).
Mechanics
2013-05-29 8:05 am
回答 (1)
2013-05-30 12:58 am
✔ 最佳答案
Q1. The marble undergoes circular motion before leaving the bowl, and projectile motion after thatLet v be the velocity of the marble when it reaches the rim
Loss in KE = Gain in PE
KE of marble at the rim = 1/2*mv^2
= 1/2*mu^2-mg(4a/3)
Magnitude of v = √[u^2-8ga/3] m/s
Let A be the angle COP, where C is a point vertically above O and P be the point at which the marble leaves the bowl
Then cos(A) = 1/3 -> A = 1.231 rad
Direction of v = 1.231 rad to horizontal
Vertical component of v = v*sin(1.231) = 2√2*v/3 m/s
Horizontal component of v = v*cos(1.231) = v/3 m/s
For the minimum value of v, the marble drops exactly at the rim of the opposite side
Horizontal distance travelled in projectile motion = 2*(2√2*a/3) = 4√2*a/3 m
Time required = t = s/v = (4√2*a/3)/(v/3) = 4√2*a/v s
Vertical displacement = vt+1/2*(-g)t^2 = 0
2√2*v/3*(4√2*a/v) - 0.5*g*(4√2*a/v)^2 = 0
16/3*a = 0.5*g*32*a^2/v^2
v^2 = 3ag
u^2-8ga/3 = 3ag
u = √[17ag/3] m/s
Q2.
Weight of P = Tension of AP(T1) + Tension of BP(T2) = mg
For the triangle ABP, AP^2 + BP^2 = 9a^2/16 + a^2 = (5a/4)^2
∠APB = π/2 rad
sin∠PAB = 4/5, sin∠PBA = 3/5
T1*sin∠PAB + T2*sin∠PBA = mg, T1*cos∠PAB - T2*cos∠PBA = 0
4/5*T1 + 3/5*T2 = mg -(1) and T1*3/5 - T2*4/5 = 0 -(2)
From(2), T1 = 4/3*T2
4/5*(4/3*T2) + 3/5*T2 = mg
25/15*T2 = mg
T2 = 3mg/5
Let the natural length of BP be L
k(a-L) = 3mg/5
a-L = 3mg/5k
L = a-3mg/5k = (5ka-3mg)/5k
收錄日期: 2021-04-16 15:53:29
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https://hk.answers.yahoo.com/question/index?qid=20130529000051KK00003