1)A particle starts from rest at position A at a time t = 0 seconds and changes
ID: 1859972 • Letter: 1
Question
1)A particle starts from rest at position A at a time t = 0 seconds and changes its speed thereafter at a constant rate of 2.3 m/s2 as it follows the horizontal path shown.
The linear distance travelled from A to B, L = 2.2 metres; after which the path follows a circular path of radius R = 1.9 metres
Determine the velocity (in m/s) of the particle as it passes point B
2) A particle starts from rest at position A at a time t = 0 seconds and changes its speed thereafter at a constant rate of 1.3 m/s2 as it follows the horizontal path shown.
The linear distance travelled from A to B, L = 2.6 metres; after which the path follows a circular path of radius R = 2.9 metres
Determine the velocity (in m/s) of the particle as it passes point C
3) A particle starts from rest at position A at a time t = 0 seconds and changes its speed thereafter at a constant rate of 0.6 m/s2 as it follows the horizontal path shown.
The linear distance travelled from A to B, L = 2.2 metres; after which the path follows a circular path of radius R = 3.0 metres
Determine the centripetal component of acceleration (in m/s2) of the particle as it passes a point a fraction after point B
Note, at this position, the velocity at B can be assumed.
4) A particle starts from rest at position A at a time t = 0 seconds and changes its speed thereafter at a constant rate of 1.2 m/s2 as it follows the horizontal path shown.
The linear distance travelled from A to B, L = 1.3 metres; after which the path follows a circular path of radius R = 1.1 metres
Determine the centripetal component of acceleration (in m/s2) of the particle as it passes point C
5) A particle starts from rest at position A at a time t = 0 seconds and changes its speed thereafter at a constant rate of 2.3 m/s2 as it follows the horizontal path shown.
The linear distance travelled from A to B, L = 2.0 metres; after which the path follows a circular path of radius R = 3.0 metres
Determine the total magnitude of acceleration (in m/s2) of the particle as it passes a point a fraction after point B
Note, at this position, the velocity at B can be assumed.
6) A particle starts from rest at position A at a time t = 0 seconds and changes its speed thereafter at a constant rate of 0.4 m/s2 as it follows the horizontal path shown.
The linear distance travelled from A to B, L = 1.6 metres; after which the path follows a circular path of radius R = 3.0 metres
Determine the total acceleration (in m/s2) of the particle as it passes point C
Explanation / Answer
1)
v^2 = u^2 + 2as
v^2 = 0 + 2*2.3*2.2
v = 3.18 m/s
w = v/R = 3.18 / 1.9 = 1.674 rad/s
alpha = a/r = 2.3 / 1.9 = 1.21 rad/s^2
w^2 = w0 ^2 + 2*alpha*theta
w^2 = 1.674^2 + 2*1.21*theta..................where theta = angle in radians between A and B which can be found from figure which is missing here.
Calculate w from above.
Then v = w*R = w*1.9
2)
v^2 = u^2 + 2as
v^2 = 0 + 2*1.3*2.6
v = 2.6 m/s
w = v/R = 2.6 / 2.9 = 0.8966 rad/s
alpha = a/r = 1.3 / 2.9 = 0.448 rad/s^2
w^2 = w0 ^2 + 2*alpha*theta
w^2 = 0.8966^2 + 2*0.448*theta..................where theta = angle in radians between A and B which can be found from figure which is missing here.
Calculate w from above.
Then v = w*R = w*2.9
3)
v^2 = u^2 + 2as
v^2 = 0 + 2*0.6*2.2
v = 1.625 m/s
w = v/R = 1.625 / 3 = 0.5416 rad/s
alpha = a/r = 0.6 / 3 = 0.2 rad/s^2
w^2 = w0 ^2 + 2*alpha*theta
w^2 = 0.5416^2 + 2*0.2*theta..................where theta = angle in radians between A and B which can be found from figure which is missing here.
Calculate w from above.
Then centripetal component of acceleration = R*w^2 = 3*(0.5416^2 + 2*0.2*theta)
4)
v^2 = u^2 + 2as
v^2 = 0 + 2*1.2*1.3
v = 1.766 m/s
w = v/R = 1.766 / 1.1 = 1.606 rad/s
alpha = a/r = 1.2 / 1.1 = 1.09 rad/s^2
w^2 = w0 ^2 + 2*alpha*theta
w^2 = 1.606^2 + 2*1.09*theta..................where theta = angle in radians between A and B which can be found from figure which is missing here.
Then centripetal component of acceleration = R*w^2 = 1.1*(1.606^2 + 2*1.09*theta)
5)
v^2 = u^2 + 2as
v^2 = 0 + 2*2.3*2
v = 3.03 m/s
w = v/R = 3.03 / 3 = 1.01 rad/s
alpha = a/r = 2.3 / 3 = 0.77 rad/s^2
w^2 = w0 ^2 + 2*alpha*theta
w^2 = 1.01^2 + 2*0.77*theta..................where theta = angle in radians between A and B which can be found from figure which is missing here.
Then centripetal component of acceleration a_c = R*w^2 = 3*(1.01^2 + 2*0.77*theta)
Tangential component of acceleration a_t = R*alpha = 3*0.77 = 2.3 m/s^2
Net acceleration = sqrt (a_c^2 + a_t^2)
a = sqrt [ (3*(1.01^2 + 2*0.77*theta))^2 + 2.3^2]..................where theta = angle in radians between A and B which can be found from figure which is missing here.
6)
v^2 = u^2 + 2as
v^2 = 0 + 2*0.4*1.6
v = 1.13 m/s
w = v/R = 1.13 / 3 = 0.377 rad/s
alpha = a/r = 0.4 / 3 = 0.13 rad/s^2
w^2 = w0 ^2 + 2*alpha*theta
w^2 = 0.377^2 + 2*0.13*theta..................where theta = angle in radians between A and B which can be found from figure which is missing here.
Then centripetal component of acceleration a_c = R*w^2 = 3*(0.377^2 + 2*0.13*theta)
Tangential component of acceleration a_t = R*alpha = 3*0.13 = 0.4 m/s^2
Net acceleration = sqrt (a_c^2 + a_t^2)
a = sqrt [ (3*(0.377^2 + 2*0.133*theta))^2 + 0.4^2]..................where theta = angle in radians between A and B which can be found from figure which is missing here.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.