{"id":33,"date":"2016-05-04T15:56:28","date_gmt":"2016-05-04T15:56:28","guid":{"rendered":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/?page_id=33"},"modified":"2016-05-04T15:56:28","modified_gmt":"2016-05-04T15:56:28","slug":"tutorial-8","status":"publish","type":"page","link":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/tutorial-8\/","title":{"rendered":"Tutorial 8"},"content":{"rendered":"

1\u00a0\u00a0\u00a0 A flywheel has a moment of inertia of 1.2 kgm2<\/sup>\u00a0. The flywheel is acted on by a torque of magnitude 0.80 N m.<\/p>\n

(a)\u00a0\u00a0 Calculate the angular acceleration produced.<\/p>\n

(b)\u00a0\u00a0 The torque acts for 5.0 s and the flywheel starts from rest. Calculate the angular velocity at the end of the 5.0 s.<\/p>\n

2\u00a0\u00a0 A mass of 0.10 kg is hung from the axle of a flywheel as shown below. The mass is released from a height of 2.0 m above ground level.<\/p>\n

The following results were obtained in the experiment:<\/p>\n

time for mass to fall to the ground t\u00a0 =\u00a0 8.0 s<\/p>\n

radius of axle\u00a0R\u00a0 =\u00a0 0.10 m.<\/p>\n

\"Capture\"<\/a><\/p>\n

(a)\u00a0By energy considerations, show that, ignoring friction,\u00a0the final speed of the flywheel is given by\u00a0\u00a0 \"Capture\"\u00a0 where I<\/em> is the moment of inertia of the flywheel.<\/p>\n

(b)\u00a0 Calculate the moment of inertia of the flywheel.<\/p>\n

3\u00a0\u00a0 A heavy drum has a moment of inertia of 2.0 kg m2.\u00a0 It is rotating freely at 10 revs-1<\/sup>\u00a0 and has a radius of 0.50 m.\u00a0 A constant frictional force of 5.0 N is then exerted at the rim of the drum.<\/p>\n

(a)\u00a0\u00a0 Calculate the time taken for the drum to come to rest.<\/p>\n

(b)\u00a0\u00a0 Calculate the angular displacement in this time.<\/p>\n

(c)\u00a0\u00a0 Hence calculate the heat generated in the braking action.<\/p>\n

4\u00a0\u00a0\u00a0 A cycle wheel is mounted so that it can rotate horizontally as shown.<\/p>\n

Data on wheel:\u00a0\u00a0\u00a0\u00a0\u00a0 radius of wheel = 0.50 m,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 mass of wheel = 2.0 kg.<\/p>\n

\"Capture\"<\/a><\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>(a)\u00a0 Calculate the moment of inertia of the wheel system.\u00a0 State any assumptions you make.<\/p>\n

(b)\u00a0 A constant driving force of 20 N is applied to the rim of the wheel.<\/p>\n

(i)\u00a0\u00a0 Calculate the magnitude of the driving torque on the wheel.<\/p>\n

(ii)\u00a0\u00a0 Calculate the angular acceleration of the wheel.<\/p>\n

(c)\u00a0 After a period of 5.0 s, calculate:<\/p>\n

(i)\u00a0 the angular displacement,<\/p>\n

(ii)\u00a0 the angular momentum of the wheel, and<\/p>\n

(iii)\u00a0 the kinetic energy of the wheel.<\/p>\n

5\u00a0\u00a0\u00a0\u00a0 A very light but strong disc is mounted on a free turning bearing as shown below.<\/p>\n

\"Capture\"<\/a><\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>A mass of 0.20 kg is placed at a radius of 0.40 m and the arrangement is set rotating at 1.0 revs-1<\/sup>\u00a0. (The moment of inertia of the disc can be considered to be negligible.)<\/p>\n

(a)\u00a0\u00a0 Calculate the angular momentum of the 0.20 kg mass.<\/p>\n

(b)\u00a0\u00a0 Calculate the kinetic energy of the mass.<\/p>\n

(c)\u00a0\u00a0 The mass is pushed quickly into a radius of 0.20 m.<\/p>\n

By applying the principle of conservation of angular momentum, calculate the new angular velocity of the mass in rads-1<\/sup>\u00a0.<\/p>\n

(d)\u00a0\u00a0 Find the new kinetic energy of the mass and account for any difference.<\/p>\n

6\u00a0\u00a0\u00a0\u00a0 A uniform metal rod has a mass, M, of 1.2 kg and a length, L, of 1.0 m.\u00a0 Clamped to each end of the rod is a mass of 0.50 kg as shown below.<\/p>\n

\"Capture\"<\/a><\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>(a)\u00a0\u00a0\u00a0 Calculate an approximate value for the moment of inertia of the complete arrangement about the central axis as shown.\u00a0 Assume that\u00a0\u00a0 Irod =\u00a0 ML\u00b2\/12 about this axis.<\/p>\n

(b)\u00a0\u00a0\u00a0 The arrangement is set rotating by a force of 10 N as shown in the diagram.\u00a0 The force acts at a tangent to the radius.<\/p>\n

(i)\u00a0\u00a0\u00a0 Calculate the applied torque.<\/p>\n

(ii)\u00a0\u00a0 Hence find the maximum angular acceleration. You may assume that the force of friction is negligible.<\/p>\n

(iii)\u00a0 Calculate the kinetic energy of the arrangement 4.0 s after it is set rotating.<\/p>\n

7\u00a0\u00a0\u00a0 An unloaded flywheel, which has a moment of inertia of 1.5 kgm\u00b2, is driven by an electric motor.\u00a0 The flywheel is rotating with a constant angular velocity of 52 rads-1<\/sup>.\u00a0 The driving torque, of 7.7 N m, supplied by the motor is now removed.<\/p>\n

How long will it take for the flywheel to come to rest. You may assume that the frictional torque remains constant?<\/p>\n

8\u00a0\u00a0\u00a0 A solid aluminium cylinder and a hollow steel cylinder have the same mass and radius.\u00a0 The two cylinders are released together at the top of a slope.<\/p>\n

(a)\u00a0\u00a0 Which of the two cylinders will reach the bottom first?<\/p>\n

(b)\u00a0\u00a0 Explain your answer to part (a).<\/p>\n

9\u00a0\u00a0\u00a0\u00a0 A solid cylinder and a hollow cylinder each having the same mass M and same outer radius R, are released at the same instant at the top of a slope 2.0 m long as shown below.\u00a0The height of the slope is 0.04 m.<\/p>\n

\"Capture\"<\/a><\/p>\n

M\u00a0 =\u00a0 10 kg,\u00a0 R\u00a0 =\u00a0 0.10 m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M\u00a0 =\u00a0 10 kg,\u00a0 R\u00a0 =\u00a0 0.10 m,\u00a0 r\u00a0 = 0.05 m<\/p>\n

I\u00a0 =\u00a0 MR\u00b2\/2 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 I\u00a0 =\u00a0 M(R\u00b2 +\u00a0 r\u00b2)\/2<\/p>\n

It is observed that one of the cylinders reaches the bottom of the slope before the other.<\/p>\n

(a)\u00a0\u00a0 Using the expressions given above, show that the moments of inertia for the cylinders are as follows:<\/p>\n

(i)\u00a0 solid cylinder;\u00a0\u00a0 I\u00a0 =\u00a0 0.05 kgm\u00b2 (ii)\u00a0 hollow cylinder;\u00a0\u00a0 I\u00a0 =\u00a0 0.0625 kgm\u00b2.<\/p>\n

(b)\u00a0 By energy considerations, show that the linear velocity of any cylinder at the bottom of the slope is given by:<\/p>\n

\"Capture\"<\/a><\/p>\n

(c)\u00a0\u00a0 Using the expression in (b) above, calculate the velocities of the two cylinders at the bottom of the slope and hence show that one of the cylinders arrives at the bottom of the slope 0.23 s ahead of the other.<\/p>\n","protected":false},"excerpt":{"rendered":"

1\u00a0\u00a0\u00a0 A flywheel has a moment of inertia of 1.2 kgm2\u00a0. The flywheel is acted on by a torque of magnitude 0.80 N m. (a)\u00a0\u00a0 Calculate the angular acceleration produced. (b)\u00a0\u00a0 The torque acts for 5.0 s and the flywheel starts from rest. Calculate the angular velocity at the end of the 5.0 s. 2\u00a0\u00a0 … Continue reading Tutorial 8<\/span> →<\/span><\/a><\/p>\n","protected":false},"author":6460,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-33","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/wp-json\/wp\/v2\/pages\/33","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/wp-json\/wp\/v2\/users\/6460"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/wp-json\/wp\/v2\/comments?post=33"}],"version-history":[{"count":1,"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/wp-json\/wp\/v2\/pages\/33\/revisions"}],"predecessor-version":[{"id":41,"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/wp-json\/wp\/v2\/pages\/33\/revisions\/41"}],"wp:attachment":[{"href":"https:\/\/blogs.glowscotland.org.uk\/gc\/advancedhigher\/wp-json\/wp\/v2\/media?parent=33"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}