Question

oant is 6.87 m/s. pin am-also not to scalc) rm sphere is placed to the top of the wall.height re is equal p height is twice the ix times the thread's length. Is, without slipping, down the ramp wall y revolutions will the sphere turn le? assume that g-9.80 m/s Refer to the diagram, which shows an overhead view of four point masses connected by two massless rods that intersect each other at right angles. 3. Ohecrhecad View)m, The masses and distances are as follows: m,-156kg m,-3.87 kg m,2.94 kg m, 4.68 kg d,-2.30 m d-1.75 m d, 0.912 m d,-3.40 m d, ITL rn This rigid structure is lying flat, initially at rest in the position shown, ona frictionless surface. It is pinned to that surface at the rods' intersection point. The structure can rotate about the (frictionless) pin, but the pin doesn't move. Let an x-y coordinate system be defined with its origin at the pin. (pin) Throughout the time interval 0srs 5.30 s, a counter-clockwise torque is applied to the structure. This torque varies with time: τ(t)-AA, B, where t is measured in seconds, A 6.87 N m/s, and B-0.654 Nm. The torque is removed (becomes zero) at 5.30 s At r 9.78 s, the pin is removed (frictionlessly-pulled straight up), so that the entire rigid structure is then free to move on the frictionless surface. Calculate the location (r- and y- coordinates) of the rods' intersection point att 12.0 s

Answer 1

$\tau (t) =At^{2/3}+B$ in Counter clockwise (CCW) direction 0 to 5.30 s

A = 6.87 N-m/s^(2/3) B = 0.654 N-m , torque is removed after 5.30 secs

at t = 9.78 s , the pivot or pin is removed , so the point of intersection is free to move

The moment of Inertia of system = 1.56 x 2.30^3 Kgm^2 + 3.87 x 1.75m^2 + 2.94 x 0.912^2 + 4.68 x 3.4^2 kgm^2

= 76.6 Kg-m^2

$\alpha (t) = (1/I)(At^{2/3}+B)$ from t=0 to t= 5.3 secs this non-uniform angular acceleration takes place.

From t = 5.3 sec to t= 9.78 secs ,the system continues with uniform angular velocity

$d\theta /dt= (1/I)\int_{0}^{t}\alpha (t)dt = (1/I)\int_{0}^{t}[At^{2/3}+B]dt$

$d\theta /dt= (1/I)\int_{0}^{t}\alpha (t)dt = (1/I)[(3A/5)t^{5/3}+Bt]$

=$d\theta /dt= (1/I)\int_{0}^{t}\alpha (t)dt = (1/76.6)[(3\times 6.87/5)(5.3^{5/3}+0.654\times 5.30]$

= 0.91 Radians/s ( ang velocity at 5.3 secs)

$\theta =\int_{0}^{5.3} \left ( 1/I \right )[(3/5)(A)(t^{5/3})+Bt]dt$

=$\theta =(1/I )[(3/5)(A)(t^{8/3})(3/8)+Bt^2/2]$

= $\theta =(1/76.6 )[(3/5)(6.87)(5.3^{8/3})(3/8)+0.654 \times (5.3^2/2)$

= 1.84 Radians (angle covered upto 5.3 secs)

from 5.3 to 9.78 secs

$\theta =1.84 Radians + (0.91 R/s ) (9.78-5.3)$

=5.9 R =338 degrees CCW or 22 deg clock wis

Let us assume that the masses system has not turned at all and calculate velocities of each of them

v1= 0.91 x 2.30 = 2.1 m/s in negative Y-direction ;corresponding momentum= 2.1 x1.56 =3.28 Kgp/s

v2= 0.91 x 1.75 = 1.6 m/s in negative X-direction ;corresponding momentum=3.87 x 1.6 = 6.19 Kgm/s

v3= 0.91 x 0.912 = 0.83 m/s in positive Y-dirction ; corresponding momentum =2.94 x 0.83=2.44kgm/s

v4= 0.91 x 3.40 = 3.1 m/s in positive x direction; corresponding momentum= 4.68 x 3.1 =14.51

net momentum in +X direction=8.32 Kgm/s

net momentum in negative Y-direction =0.84 Kgm/s

net velocity with Resp to X-direction=8.32/(1.56+3.87+2.94+4.68)=0.64m/s

net velocity in negative Y direction=0.84/(1.56+3.87+2.94+4.68)=0.06 m/s

from 9.78 secs to 12.0 secs

distance moved by pin in X axis = (12.0-9.78) x0.64m/s = 1.42 m

distance moved by pin along Negative Y-axis = -0.08m