Cross section of a 1/4 of a cylindrical shape is drawn below with radius R = 0.8
ID: 1433219 • Letter: C
Question
Cross section of a 1/4 of a cylindrical shape is drawn below with radius R = 0.8 m and with the lower end on the ground and tangent to it. A block with mass m = 2.0 kg is released from the top with no initial velocity and it slides down with no friction. If C is the center of the cylinder and theta is the angular location of the block, when theta = 40 degree determine velocity of the block. the value of the normal force from the surface. Tangential, centripetal, and total accelerations. If a spring with stiffness k = 1500 N/m is placed on the horizanthal surface at point D, how much would it be compressed?Explanation / Answer
From the figure,
V = (R*g)(1/2)
V = (0.8*9.8) (1/2)
V = 2.8 m/s……………………….equation 1.
Therefore we can write
T = mV2/r
Normal to the surface can be calculated,
N + tension + centrepetal force = mgcos()
Therefore,
N = mgcos() -tension - centrepetal force
N = mgcos() – 2 *centrepetal force
N = 2 * 9.8 * cos (40) – 2* 19.6
N = 19.6* 0.7660 – 39.2
N = 15.0136 – 39.2 = -4.58
In magnitude,
It is N = 4.58 newton
c) Now we can calculate the centripetal force,
F(centripetal) = mV2/r,
F(centripetal) = 2 *(2.8)2/0.8
F(centripetal) = 15.68/0.8 = 19.6 N………………………equation 2.
Now centrepetal accleration is,
A (centrepetal) = V2/r,
A(centripetal) = (2.8)2/0.8
A(centripetal) = 9.8 m/s2……………………………………….equation3.
Tangential accleration is defined
A(Tangential) = dV/dt = 0…………………………………equation 4
Therefore, Total accleration is,
A (Total) = (A(centripetal)2 + A(Tangential)2)(1/2)
A (Total) = A(centripetal) = 9.8 m/s2 ………….from equation 3.
d) Now k = 1500 N/m
According to Hooke’s law
F = -Kx…………where negative direction shows comprssion
Therefore,
X = F/k = (m*A)/K
X= (2 *9.8)/1500
X = 1.3 *10-2 m
It is the compressed distance.
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